USACO money packageDP

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裸0/1背包，就是從各種幣種裡面拿來湊足N元，求最多有多種方案。用dp[i][j]表示選前i個幣種湊成j的方案數量

狀態轉移方程： dp[i][j] = dp[i- 1][j]    j < coins[i]      dp[i][j] = dp[i-1][j] + dp[i-1][j-coins[i]]  j >= coins[i]

/*ID:kevin_s1PROG:moneyLANG:C++*/#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <map>#include <set>#include <algorithm>#include <cstdlib>#include <list>#include <cmath>using namespace std;#define MAXN 10010//gobal variable====int V;long long N;int coins[30];long long dp[30][MAXN];//==================//function==========//==================int main(){freopen("money.in","r",stdin);freopen("money.out","w",stdout);cin>>V>>N;int coin;for(int i = 1; i <= V; i++){cin>>coin;coins[i] = coin;}for(int i = 1; i <= V; i++)dp[0][i] = 0;dp[0][0] = 1;for(int i = 1; i <= V; i++){for(long long j = 0; j < coins[i]; j++){dp[i][j] = dp[i - 1][j];}for(long long j = coins[i]; j <= N; j++){dp[i][j] = dp[i - 1][j] + dp[i][j - coins[i]];}}cout<<dp[V][N]<<endl;return 0;}