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HDU's ACM 2110 Crisis of HDU

2014-11-17 23:34 375 查看
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 105

int 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]);
else
printf("sorry\n");
}
return 0;
}</span>
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标签:  杭电 acm 算法 母函数