ACM 442 - Matrix Chain Multiplication

#include <stdio.h>
#include <ctype.h>

struct matrix
{
    char ch;
    int row;
    int col;
    int count;
};
struct matrix m[26], stack[200];

int main()
{
    int n, row, col, i;
    char ch, str[200];
    scanf("%d", &n);
    getchar();
    for (i = 0; i < n; i ++)
    {
        scanf("%c %d %d", &ch, &row, &col);
        getchar();
        int index = ch - 'A'; 
        m[index].ch = ch, m[index].row = row, m[index].col = col; 
    }
    while (gets(str))
    {
        int count = 0, error = 0, index = 0;
        int row = 0, col = 0;
        for (i = 0; (ch = str[i]); i ++)
        {
            if (ch == '(') stack[index ++].ch = ch;
            if (isupper(ch))
            {
                int s = ch - 'A';
                row = m[s].row, col = m[s].col;
                if (isupper(stack[index - 1].ch))
                {
                    if (stack[index - 1].col != row)
                    { error = 1; break; }
                    stack[index - 1].count += stack[index - 1].row * row * col;
                    stack[index - 1].col = col; 
                }
                else
                {
                    stack[index].ch = ch;
                    stack[index].row = row;
                    stack[index].col = col;
                    stack[index].count = 0;
                    index ++;  
                }
            }
            if (ch == ')') 
            {
                if (index - 1 < 0) { error = 1; break; }
                if (stack[index - 1].ch == '(') index --;
                if (index - 2 >= 0 && 
                    isupper(stack[index - 1].ch) && stack[index - 2].ch == '(')
                {
                    stack[index - 2].ch = stack[index - 1].ch;
                    stack[index - 2].row = stack[index - 1].row;
                    stack[index - 2].col = stack[index - 1].col;
                    stack[index - 2].count = stack[index - 1].count;
                    index --;
                    if (isupper(stack[index - 2].ch))
                    {
                        row = stack[index - 1].row;
                        col = stack[index - 1].col;
                        count = stack[index - 1].count;
                        
                        if (stack[index - 2].col != row)
                        { error = 1; break; }
                        stack[index - 2].count += (count + stack[index - 2].row * row * col);
                        stack[index - 2].col = col;
                        index --;
                    }
                }
            }
                
        }
        
        if (error == 0) printf("%d\n", stack[0].count);
        else if (error) printf("error\n");
        
    }
    return 0;
}

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ACM 424 - Integer Inquiry

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define NUMLEN 200

struct AddNum
{
    int num[NUMLEN];
    int length;
};

struct AddNum AN1, AN2;
char numStr[101];

int main()
{
    int i, j;
    while (gets(numStr))
    {
        if (numStr[0] == '0') break;
        AN2.length = strlen(numStr) - 1;
        for (i = AN2.length, j = 0; i >= 0; i --, j ++)
            AN2.num[j] = numStr[i] - '0'; 
        for (i = 0; i < NUMLEN; i ++)
        {
            AN1.num[i] += AN2.num[i];
            if (AN1.num[i] >= 10)
               AN1.num[i + 1] ++, AN1.num[i] %= 10;
        }
    }
    for (i = NUMLEN - 1; ; i --)
        if (AN1.num[i] > 0)
           break;
    for (; i >= 0; i --)
        printf("%d", AN1.num[i]);
    printf("\n");   
    return 0;
}

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ACM 375 - Inscribed Circles and Isosceles Triangles

#include <stdio.h>
#include <math.h>
double B, H, SL, S, area, r, HSub2R, cycleLen;

int main()
{
    double PI = 2*acos(0.0);
    int n, i;
    scanf("%d", &n);
    for (i = 0; i < n; i ++)
    {
        cycleLen = 1e-12;
        scanf("%lf %lf", &B, &H);
        B += 1e-15; H += 1e-15; 
        SL = (double)sqrt( (H*H) + (B*B/4) );
        area = B * H / 2;
        r = 2 * area / (SL+SL+B);
        while (r >= 0.000001)
        {
            cycleLen += 2 * r * PI;  
            HSub2R = H - 2 * r;
            r *= HSub2R / H;
            H = HSub2R;
        }
        if (i != 0) printf("\n");
        printf("%13.6lf\n", cycleLen);
    }
    return 0;
}

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ACM 374 - Big Mod

#include <stdio.h>
int f(int m,int n, int mod)
{
  long long int mul = m;
  long long int r = 1;
  while(n){
      if(n % 2 == 1){r *= mul;r %= mod;}
      mul *= mul;
      mul %= mod;
      n >>= 1;
  }
  return r;
}
int main()
{
    int B,P,M;
    while(scanf("%d%d%d",&B,&P,&M)==3)
      printf("%d\n",f(B,P,M));
    return 0;
}

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ACM 344 - Roman Digititis

#include <stdio.h>

struct RomanNum
{
   int i;
   int v;
   int x;
   int l;
   int c;       
};

struct RomanNum RN[101];

void adder(int index, int n)
{
     if (n == 0) return;
     if (n >= 40 && n <= 49)
         RN[index].l ++, RN[index].x ++, n -= 40;     
     if (n >= 90 && n <= 99)
         RN[index].c ++, RN[index].x ++, n -= 90;
     if (n % 10 == 4)
         RN[index].v ++, RN[index].i ++, n -= 4;     
     if (n % 10 == 9)
         RN[index].x ++, RN[index].i ++, n -= 9;
     if (n / 100 > 0)
         RN[index].c += (n/100), n %= 100;
     if (n / 50 > 0)
         RN[index].l += (n/50), n %= 50;
     if (n / 10 > 0)
         RN[index].x += (n/10), n %= 10;
     if (n / 5 > 0)
         RN[index].v += (n/5), n %= 5;
     if (n / 1 > 0)
         RN[index].i += (n/1), n %= 1;                    
}

int main()
{
    int i, n;
    for (i = 0; i < 100; i ++)
    {
        adder(i, i + 1);
        RN[i + 1].i = RN[i].i, RN[i + 1].v = RN[i].v, RN[i + 1].x = RN[i].x;
        RN[i + 1].l = RN[i].l, RN[i + 1].c = RN[i].c;  
    }
    
    while (1)
    {
        scanf("%d", &n);
        if (n == 0) break;
        n -= 1;
        printf("%d: %d i, %d v, %d x, %d l, %d c\n", n+1, RN[n].i, RN[n].v, RN[n].x, RN[n].l, RN[n].c);  
    }
    return 0;
}

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ACM 340 - Master-Mind Hints

#include <stdio.h>

struct Ans
{
    int ansInt;
    int isComp;   
};

int main()
{
    int n, i, m, isZero, A, B, j, index = 0;
    while (scanf("%d", &n) == 1)
    {
        if (n == 0) break; 
        printf("Game %d:\n", ++ index); 
        struct Ans a[n];
        for (i = 0; i < n; i ++)
        {
            scanf("%d", &m);
            a[i].ansInt = m;
            a[i].isComp = 0;
        }
        struct Ans b[n];
        while (1)
        {
            isZero = 1, A = 0, B = 0;
            for (i = 0; i < n; i ++)
            {
                scanf("%d", &m);
                if (m != 0) isZero = 0;
                b[i].ansInt = m;
                b[i].isComp = 0;
                a[i].isComp = 0;
            }
            for (i = 0; i < n; i ++)
                if (a[i].ansInt == b[i].ansInt)
                   A ++, a[i].isComp = 1, b[i].isComp = 1;
            for (i = 0; i < n; i ++)
                for (j = 0; j < n; j ++)
                {
                    if (i != j && a[i].isComp == 0 && b[j].isComp == 0 && a[i].ansInt == b[j].ansInt)
                       B ++, a[i].isComp = 1, b[j].isComp = 1;
                }          
            if (isZero) break;
            else printf("    (%d,%d)\n", A, B);
        }
           
    }
    return 0;
}

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ACM 350 - Pseudo-Random Numbers

#include <stdio.h>

int PseRand(int Z, int I, int M, int L)
{
    return (Z * L + I) % M;
}

int main()
{
    int Z, I, M, L, caseNum = 1;
    
    while (1)
    {
        scanf("%d %d %d %d", &Z, &I, &M, &L);
        if (Z == 0 && I == 0 && M == 0 && L == 0) break;
        int isTrue[10000] = {0}, len[10000] ;
        printf("Case %d: ", caseNum++);
        int count = 0;
        while (!isTrue[L])
        {
            isTrue[L] = 1, len[L] = count ++;  
            L = PseRand(Z, I, M, L);  
        }
        printf("%d\n", count - len[L]);
    }
    
    return 0;
}

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ACM 299 - Train Swapping

#include <stdio.h>
#include <malloc.h>
int *S, size;

int bubbleSort ()
{
    int i, j, tmp, count = 0;
    for (i = 0; i < size; i ++)
    {
        for (j = 0; j < size - i - 1; j ++)
        {
            if (S[j] > S[j + 1])
            {
                 tmp = S[j];
                 S[j] = S[j + 1];
                 S[j + 1] = tmp; 
                 count ++;    
            }
        }
    }
    return count;
}

int main()
{
    /*freopen("111.txt", "r", stdin);
    freopen("111w.txt", "w", stdout);*/
    int n, i, j;
    scanf("%d", &n);
    for (i = 0; i < n; i ++)
    {
        scanf("%d", &size);
        
        S = (int *)malloc(sizeof(int)*size);
        for (j = 0; j < size; j ++)
            scanf("%d", &S[j]);
        printf("Optimal train swapping takes %d swaps.\n", bubbleSort());
        free(S);
    }    
}

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ACM 333 - Recognizing Good ISBNs

#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main()
{
    char str[82];
    int i, numCount, s, e;
    while (gets(str))
    {
        if (strlen(str) == 0) { printf(" is incorrect.\n"); continue; }
        numCount = 0;
        int per = 0;
      int tsum = 0;
        int sum[9] = {0}, wrong = 0, d = 0, isX = 0, g = 0;
        for (s = 0; str[s] == ' ' || str[s] == '\t' || isalpha(str[s]) ; s ++) 
            if (str[s] == 'X') break; 
        for (e = strlen(str) - 1; str[e] == ' ' || str[e] == '\t' || isalpha(str[e]); e --)
            if (str[e] == 'X') break;  
        for (i = s; i <= e; i ++)
        {
            if (str[i] == '-') continue;
            else if (isdigit(str[i]))
            {
                numCount ++; 
                per += str[i] - '0';
          tsum += per;
            }
            else if (str[i] == 'X')
            {
                if (numCount != 9) { wrong = 1; break;}
                else
                {
                    per += 10;
              tsum += per;
                    numCount ++;
                }
            }
            else { wrong = 1; break;}
        }
        for (i = 0; str[i] == ' ' || str[i] == '\t'; i ++) ;
        s = i;
        for (i = strlen(str) - 1; str[i] == ' ' || str[i] == '\t'; i --) ;
        e = i;
        for (i = s; i <= e; i ++)
            putchar(str[i]);
        if (wrong) printf(" is incorrect.\n");
        else if (numCount >= 10 && tsum % 11 == 0) printf(" is correct.\n");
        else  printf(" is incorrect.\n");
                
    }
    return 0;
}

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ACM 332 - Rational Numbers from Repeating Fractions

#include <stdio.h>
#include <string.h>

int gcd(int a,int b)                                                                    
{                                                                                                 
    if (a % b == 0){ return b;}                                                                       
  return gcd(b , a % b);                                                                            
}

int pow10(int n)
{
    int i, sum = 1;
    for (i = 0; i < n; i ++)
        sum *= 10;
    return sum;    
}

char str[20];

int main()
{
    int j, i, k, h, g, d, s, caseNum = 1;
    while(scanf("%d", &j) == 1 && j != -1)
    {
        scanf("%s", str);            
        char ch;
        int len = strlen(str);
        if (j != 0) d = pow10(len - 2) - pow10(len - 2 - j);
        else d = pow10(len - 2);
        k = 0, h = 0;
        for (i = 2, g = 0; i < len; i ++, g ++)
        {
            ch = str[i];
            if (g < (len - 2 - j)) h = h * 10 + (ch - '0');
            k = k * 10 + (ch - '0');
        }
        if (j != 0)
        { 
           s = gcd(k - h, d);
           printf("Case %d: %d/%d\n", caseNum ++, (k - h) / s, d / s);
        }
        else
        { 
            s = gcd(k, d);
            printf("Case %d: %d/%d\n", caseNum ++, k / s, d / s);
        }
       
        
    }
    
    return 0;
}

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ACM 326 - Extrapolation Using a Difference Table

#include <stdio.h>

int a[50];

int main() {
    int n, i, j, k;
    scanf("%d", &n);
    while (n) 
    {
        for (i=0; i<n; i++)
            scanf("%d", &a[i]);
        scanf("%d", &k);
        for (j=n-1; j>0; j--)
            for (i=0; i<j; i++)
                a[i] = a[i+1] - a[i];
        for (i=0; i<k; i++)
            for (j=1; j<n; j++)
                a[j] += a[j-1];
        printf("Term %d of the sequence is %d\n", n+k, a[n-1]);
        scanf("%d", &n);
    }
    return 0;
}

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ACM 311 - Packets

#include <stdio.h>

int residual[6], packets[6];

int main()
{
    while (1)
    {   
        int i, isZero = 1, total = 0;
        for (i = 0; i < 6; i ++)
        {
            scanf("%d", &packets[i]);
            residual[i] = 0;
            if (packets[i] != 0)
               isZero = 0;
        }
        if (isZero) break;
        total += packets[5]; /* 6*6一人一個盒子 */
        total += packets[4]; /* 5*5一人一個盒子，剩餘 11 個 1*1 空間 */ 
        residual[0] += packets[4] * 11;
        total += packets[3]; /* 4*4一人一個盒子，剩餘 5 個 2*2 空間 */
        residual[1] += packets[3] * 5;
        
        total += packets[2] / 4; /* 3*3 之處理方式 */
        packets[2] %= 4, residual[2] = packets[2];
        if (residual[2] == 1)
           residual[2] = 0, residual[0] += 7, residual[1] += 5, total ++;
        else if (residual[2] == 2)
           residual[2] = 0, residual[0] += 6, residual[1] += 3, total ++;
        else if (residual[2] == 3)
           residual[2] = 0, residual[0] += 5, residual[1] += 1, total ++;
           
        if (residual[1] >= packets[1]) /* 2*2 之處理方式 */
           residual[1] -= packets[1], packets[1] = 0;
        else
           packets[1] -= residual[1], residual[1] = 0;
        total += packets[1] / 9;
        packets[1] %= 9;
        if (packets[1] > 0)
           residual[0] += 36 - packets[1] * 4, packets[1] = 0, total ++; 
        else residual[0] += residual[1] * 4, residual[1] = 0;
         
        if (residual[0] >= packets[0]) /* 1*1 之處理方式 */
           residual[0] -= packets[0], packets[0] = 0;
        else
           packets[0] -= residual[0], residual[0] = 0;
        total += packets[0] / 36;
        packets[0] %= 36;
        if (packets[0] > 0)
           total ++;    
        printf("%d\n", total);   
    }
    return 0;
}

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ACM 300 - Maya Calendar

#include <stdio.h>
#include <string.h>

char hMonthStr[19][8] = {"pop", "no", "zip", "zotz", 
                         "tzec", "xul", "yoxkin", "mol", 
                         "chen", "yax", "zac", "ceh", 
                         "mac", "kankin", "muan", "pax", 
                         "koyab", "cumhu", "uayet"};
char tMonthStr[20][10] = {"imix", "ik", "akbal", "kan",
                         "chicchan", "cimi", "manik", "lamat", 
                         "muluk", "ok", "chuen", "eb", 
                         "ben", "ix", "mem", "cib", 
                         "caban", "eznab", "canac", "ahau"};
char dStr[6], mStr[8];
int main()
{
    int hDay, hMonth, hYear, tDay, tMonth, tYear;
    int n, days, i;
    scanf("%d", &n);
    printf("%d\n", n);
    while (n --)
    {
        char *d;
        scanf("%s %s %d", dStr, mStr, &hYear);
        d = strtok(dStr, ".");
        sscanf(d, "%d", &hDay);
        for (i = 0; i < 19; i ++)
        {
            if (strcmp(hMonthStr[i], mStr) == 0)
            { hMonth = i; break; }
        }
        days = hDay + hMonth * 20 + hYear * 365;
        tYear = days / 260;
        days %= 260;
        tMonth = days % 20;
        tDay = days % 13 + 1;
        
        printf("%d %s %d\n", tDay, tMonthStr[tMonth], tYear);
    }
    return 0;
}

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ACM 272 - TEX Quotes

#include <stdio.h>

int main()
{
    int style = 0;
    char c;
    while (1){
        c = getchar();
        if (c == EOF) break;
        if (c == '\"')
        {
            if (style == 0)
            {
               printf("``");
               style = 1;
            }
            else
            {
               printf("''");
               style = 0;
            }
        }
        else putchar(c);
    }
    return 0; 
}

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ACM 264 - Count on Cantor

#include<stdio.h>    
#include<stdlib.h>   
#include<string.h>   
#include<math.h>   
main()   
{   
    int n,a,b,c,d;   
    while(scanf("%d",&n)==1)   
    {   
        for(d=1,c=n;c>d;d++) c=c-d;   
        if ( d%2==1 ) a=d-c+1;   
        else a=c;   
        b=d+1-a;   
        printf("TERM %d IS %d/%d\n",n,a,b);   
    }   
    return 0;   
}

以上程式轉自：http://mypaper.pchome.com.tw/iustlovefish/post/1311842753
稍微說明一下程式碼，先看以下圖形：標紅色部分，依照每斜行遞增，所以 d=1, c=n 初始化後，用一迴圈每做一次 c 減掉 d 而做完後 d 累加一，所以這一部份是做算出 n 是在第 d 斜行的第 c 個位置。
而這個圖表有一個特性，請看以下表：

x - 1 / y - 1	x - 1 / y	x - 1 / y + 1
x / y - 1	x / y	x / y + 1
x + 1 / y - 1	x + 1 / y	x + 1 / y + 1

只要 d 在奇數它是往左下走，偶數則往右上走，而只要 d 在奇數則該斜行的第一個位置為 1 / d，偶數為 d / 1，最後再依照它的位置算出正確答案。

此提與 ACM 10182 Bee Maja 頗像，我是用圖法煉鋼的方法寫出，但似乎比上面的程式碼還慢，所以還是參考他的做法。

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ACM 256 - Quirksome Squares

#include <stdio.h>

int main( void )
{
    int i, n, j;
    while (1)
    {
        if (scanf("%d", &n) < 1) break;
        if (n == 2)
        {  
            for (i = 0; i < 10; i ++)
                for (j = 0; j < 10; j ++)
                {
                    if ( i + j >= 10) break;
                    if ((i + j)*(i + j) == (i*10 + j))
                      printf("%02d\n",(i*10 + j));
                }
        }
        if (n == 4)
        {  
            for (i = 0; i < 100; i ++)
                for (j = 0; j < 100; j ++)
                {
                    if ( i + j >= 100) break;
                    if ((i + j)*(i + j) == (i*100 + j))
                      printf("%04d\n",(i*100 + j));
                }
        }
        if (n == 6)
        {  
            for (i = 0; i < 1000; i ++)
                for (j = 0; j < 1000; j ++)
                {
                    if ( i + j >= 1000) break;
                    if ((i + j)*(i + j) == (i*1000 + j))
                      printf("%06d\n",(i*1000 + j));
                }
        }
        if (n == 8)
        {  
            for (i = 0; i < 10000; i ++)
                for (j = 0; j < 10000; j ++)
                {
                    if ( i + j >= 10000) break;
                    if ((i + j)*(i + j) == (i*10000 + j))
                      printf("%08d\n",(i*10000 + j));
                }
        }
    }
 return 0;
}

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ACM 253 - Cube painting

#include <stdio.h>

char str[15] = {"\0"};
struct Cube
{
    char ch1;
    char ch2;
    int isComp;
};

struct Cube c[6]; 

int main()
{
    while (gets(str))
    {
        int isTurn = 0, i, j;
        c[0].ch1 = str[0], c[0].ch2 = str[5], c[0].isComp = 0;
        c[1].ch1 = str[1], c[1].ch2 = str[4], c[1].isComp = 0;
        c[2].ch1 = str[2], c[2].ch2 = str[3], c[2].isComp = 0;
        c[3].ch1 = str[6], c[3].ch2 = str[11], c[3].isComp = 0;
        c[4].ch1 = str[7], c[4].ch2 = str[10], c[4].isComp = 0;
        c[5].ch1 = str[8], c[5].ch2 = str[9], c[5].isComp = 0;
        for (i = 0; i < 3; i ++)
            for (j = 3; j < 6; j ++)
            {
                if (c[i].isComp == 0 && c[j].isComp == 0)
                {
                    if (c[i].ch1 == c[j].ch1 && c[i].ch2 == c[j].ch2 ||
                        c[i].ch1 == c[j].ch2 && c[i].ch2 == c[j].ch1)
                    {
                        c[i].isComp = 1, c[j].isComp = 1;
                        break;
                    }
                }
            }
        for (i = 0; i < 6; i ++)
            if (c[i].isComp == 0)
               isTurn = 1;
        if (isTurn)
            printf("FALSE\n");
        else printf("TRUE\n");                     
    }
    return 0;
}

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ACM 202 - Repeating Decimals

#include <stdio.h>
#include <memory.h>

#define MAX 5000
#define DISPLAY_LIMIT 50
#define MAX_INT 3000

int min (int a , int b){
    return a > b ? b : a;
}


int main()
{
    int digits[MAX + 1], remainderExist[MAX_INT], remainderPos[MAX_INT];
    while (1)
    {
        int numerator, denominator, oNumerator, quotient, remainder, recode, i;
        
        if (scanf("%d", &numerator) < 1) break;
        oNumerator = numerator;
        scanf("%d", &denominator);
        
        memset (remainderExist, 0, sizeof(remainderExist));
        memset (remainderPos, 0, sizeof(remainderPos));
        
        quotient = numerator / denominator;
        remainder = numerator % denominator;
        
        recode = quotient;
        int n = 0, isCycle = 0, cyclePos = MAX, cycleLen = 0;
        
        while ( n <= MAX && !isCycle)
        {
            if (remainderExist[remainder] )
            {
                cyclePos = remainderPos[remainder];
                cycleLen = n - cyclePos;
                isCycle = 1;
            }
            else
            {
                remainderExist[remainder] = 1;
                remainderPos[remainder] = n;
            }
            numerator = remainder * 10;
            
            quotient = numerator / denominator;
            remainder = numerator % denominator;
            digits[n] = quotient;
            n ++;
        }
        printf("%d/%d = %d.", oNumerator, denominator, recode);
        int limit = min(cyclePos, DISPLAY_LIMIT);
        
        for (i = 0; i < limit; ++i )
            printf("%d", digits[i]);
        
        if (cyclePos < DISPLAY_LIMIT)
        {
            printf("(");
            limit = min(n - 1, DISPLAY_LIMIT);
            for (i = cyclePos; i < limit; ++i ) 
                printf("%d", digits[i]);
            if (n > DISPLAY_LIMIT)
                printf("...");
            printf(")");         
        }    
        printf("\n");
        printf("   %d = number of digits in repeating cycle\n\n",cycleLen );

    }
    return 0;
}

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ACM 11799 - Horror Dash

#include <stdio.h>

int getInt()
{
    char ch;
    int n = 0;
    while( ch = getchar())
        if(ch != ' ' && ch != '\n') break;
    n = ch - 48;
    while( ch = getchar())
    {
        if(ch == ' ' || ch == '\n') break;
        n = n * 10 + ch - 48;
    }
    return n;
}

int main()
{
    int caseNum, i, n, m, max, j;
    caseNum = getInt();
    
    for (i = 1; i <= caseNum; i ++)
    {
        n = getInt();
        max = 0;
        while (n --)
        {
            m = getInt();
            if (m > max) max = m;
        }
        printf("Case %d: %d\n", i, max);
    }
    return 0;
}

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ACM 11805 - Bafana Bafana

#include <stdio.h>

int getInt()
{
    char ch;
    int n = 0;
    while( ch = getchar())
        if(ch != ' ' && ch != '\n') break;
    n = ch - 48;
    while( ch = getchar())
    {
        if(ch == ' ' || ch == '\n') break;
        n = n * 10 + ch - 48;
    }
    return n;
}

int main()
{
    int i, caseNum, N, K, P, j;
    caseNum = getInt();
    for (i = 1; i <= caseNum; i ++)
    {
        N = getInt(), K = getInt(), P = getInt();
        j = (K + P) % N;
        if (j == 0) j = N;
        printf("Case %d: %d\n", i, j);
    }
    return 0;
}

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