HDU's ACM 2152 Fruit

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原题链接：HDU's ACM 2152 Fruit

分析：母函数大法好！参考：HDU's ACM 1028 Ignatius and the Princess III，变化就在于范围，设某种水果的数量范围为 [Min, Max]，则对应的因式为(x^Min+x^(Min+1)+...+x^Max)，无其他难点。

AC Code

<span style="font-family:Microsoft YaHei;font-size:14px;">#include <stdio.h>#include <string.h>#define MAXN 100#define MAXM 100+10typedef struct {int min;int max;} Range;Range rangeArr[MAXN];int coeff1[MAXM];int coeff2[MAXM];int main(){int n, m;int i, j, k;while(scanf("%d%d", &n, &m) == 2){memset(rangeArr, 0, sizeof(int)*MAXN);memset(coeff1, 0, sizeof(int)*MAXM);memset(coeff2, 0, sizeof(int)*MAXM);for(i=0;i<n;++i){scanf("%d%d", &rangeArr[i].min, &rangeArr[i].max);}for(i=rangeArr[0].min;i<=rangeArr[0].max && i<=m;++i)coeff1[i] = 1;for(i=1;i<n;++i){for(j=0;j<=m;++j)for(k=rangeArr[i].min;k<=rangeArr[i].max && (j+k)<=m;++k)coeff2[j+k] += coeff1[j];memcpy(coeff1, coeff2, sizeof(int)*MAXM);memset(coeff2, 0, sizeof(int)*MAXM);}printf("%d\n", coeff1[m]);}return 0;}</span>

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