Coin Combinations II(Dynamic Programming)

题目描述

Consider a money system consisting of n coins. Each coin has a positive integer value. Your task is to calculate the number of distinct ordered ways you can produce a money sum x using the available coins.
For example, if the coins are {2,3,5} and the desired sum is 9, there are 3 ways:

2+2+5
3+3+3
2+2+2+3

输入

The first input line has two integers n and x: the number of coins and the desired sum of money.
The second line has n distinct integers c1,c2,...,cn: the value of each coin.
Constraints
1 ≤ n ≤ 100
1 ≤ x ≤ 
1 ≤ ci ≤ 

输出

Print one integer: the number of ways modulo .

样例输入

3 9
2 3 5

样例输出

3

思路分析

与题目Coin Combinations II(Dynamic Programming)稍有不同。

外层循环遍历硬币面额，对于每一个硬币面额 j。

内层循环从该硬币面额 j 开始遍历到目标金额 x，对于每个目标金额 i，更新 dp[i] 的值。状态转移方程为 dp[i] += dp[i - j]，表示使用当前硬币面额 j 可以凑出金额 i 的组合数等于不使用该硬币时凑出金额 i 的组合数加上使用该硬币时凑出金额 i - j 的组合数。

每次更新后，对 dp[i] 取模，避免溢出。

代码

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll N=1e6+1;
const ll mod=1e9+7;
ll n,x;
int main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);cin>>n>>x;vector<ll>a(n,0);for(ll i=0;i<n;i++){cin>>a[i];}sort(a.begin(),a.end());vector<ll>dp(N,0);dp[0]=1;for(ll&j:a){for(ll i=j;i<=x;i++){dp[i]+=dp[i-j];dp[i]%=mod;}}cout<<dp[x];return 0;
}