AcWing 1023. 买书 完全背包

//完全背包  求方案数目
//f[i][j]   只从前i个物品中选，且总体积恰好为j的方案的集合
//f[i][j]=f[i-1][j]+f[i-1][j-v*1]+f[i-1][j-v*2]+...f[i-1][j-v*s] 
//f[i][j-v]=        f[i-1][j-v*1]+f[i-1][j-v*2]+...f[i-1][j-v*s]
//所以  f[i][j]=f[i-1][j]+f[i][j-v] 
#include <iostream>
using namespace std;
const int N = 1010;
int n;
int v[4] = {10, 20, 50, 100};
int f[N];
int main() {
    cin >> n;
    f[0] = 1;
    for (int i = 0; i < 4; i ++ )
        for (int j = v[i]; j <= n; j ++ )
            f[j] =f[j] + f[j - v[i]];
    cout << f[n] << endl;
    return 0;
}