HDU's ACM 2110 Crisis of HDU

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原题链接：HDU's ACM 2110 Crisis of HDU

分析：又一道母函数题目，可参考HDU's ACM 1028 Ignatius and the Princess III，思路类似，不同之处在于系数与指数由pi,mi指定。

AC Code:

<span style="font-family:Microsoft YaHei;font-size:14px;">#include <stdio.h>#include <string.h>#define MAXN 10000#define NUM_MAX 105int coeff1[MAXN];int coeff2[MAXN];int p[NUM_MAX];int m[NUM_MAX];int main(){int n;int i, j, k, end;int sum, target;while(scanf("%d", &n) == 1 && n) {sum = 0;for(i=0;i<n;++i) {scanf("%d%d", &p[i], &m[i]);sum += p[i]*m[i];}if(sum%3){printf("sorry\n");continue;}target = sum/3;memset(coeff1, 0, sizeof(coeff1));memset(coeff2, 0, sizeof(coeff2));end = p[0]*m[0];for(i=0;i<=end && i<=target;i+=p[0])coeff1[i] = 1;for(i=1;i<n;++i){end = p[i]*m[i];for(j=0;j<=target;++j)for(k=0;k<=end;k+=p[i]){if(j+k>target)break;coeff2[j+k] += coeff1[j];}for(j=0;j<=target;++j){coeff1[j] = coeff2[j]%10000;coeff2[j] = 0;}}if(coeff1[target])printf("%d\n", coeff1[target]);elseprintf("sorry\n");}return 0;}</span>

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