usaco: Money Systems

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res(i, j)表示用前 j 种组合出 i 。那么：res(i, j) = res(i - coin[i], j - 1) + res(i, j - 1)，即结果有两种情况：取第 j 种钱和不取第 j 种钱。

/*ID: LANG: C++TASK: money*/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#define IN "money.in"#define OUT "money.out"#define MaxV 30#define MaxN 10005using namespace std;int v, n;int coin[MaxV];long long res[MaxN][MaxV];  //res(i, j): 用前j种钱组合出ivoid solve(){    memset(res, 0, sizeof(0));    int tmp = coin[0];    while(tmp <= n) {        res[tmp][1] = 1;        tmp += coin[0];    }    for(int i = 1; i <= v; i++)        res[0][i] = 1;        for(int i = 2; i <= v; i++) {        for(int j = 1; j <= n; j++) {            if(j - coin[i-1] >= 0)                res[j][i] = res[j-coin[i-1]][i] + res[j][i-1];            else                res[j][i] = res[j][i-1];        }    }}int main(){    freopen(IN, "rb", stdin);    freopen(OUT, "wb", stdout);    while(scanf("%d%d", &v, &n) != EOF) {        for(int i = 0; i < v; i++)            scanf("%d", &coin[i]);                solve();                printf("%lld\n", res[n][v]);    }        fclose(stdin);    fclose(stdout);       return 0;}