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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int n, m;
int f[N][N];
int main() {
//文件输入
freopen("ZhiDuo_WQ.in", "r", stdin);
scanf("%d %d", &n, &m);
for (int i = 0; i <= m; i++) f[0][i] = 1; //前0种物品中选择最多不超过m的体积下选择的方案只有一种合法就是啥也不选
for (int i = 1; i <= n; i++) {
int v;
scanf("%d", &v);
/*写法1*/
for (int j = 0; j <= m; j++) {
if (j >= v)
f[i][j] = f[i - 1][j] + f[i][j - v];
//注意这里与01背包的差别
// 在01背包的状态转移中i就是i,它之前不能出现过i,只能从i-1转移
// 在完全背包的状态转移中i出现过没关系只要能装下出现几个都行可以从i转移
// 01背包
// f[i][j] = f[i - 1][j] + f[i - 1][j - v];
else
f[i][j] = f[i - 1][j];
}
/*写法2*/
// for (int j = 0; j <= m; j++) {
// f[i][j] = f[i - 1][j];
// if (j >= v) f[i][j] += f[i][j - v];
// }
}
printf("%d\n", f[n][m]);
return 0;
}