hackerrank The Coin Change Problem(dp)
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题目:https://www.hackerrank.com/challenges/coin-change/problem
题意:给你m种硬币,每种数量不限,问你硬币的和为n的方案数
思路:开始想的用dp[i][j]表示i种硬币组成和为j的方案数,那么dp[i][j] = dp[i-1][j] + dp[i][j-a[i]];
ac之后发现可以缩成一维(开始就该想到的,不熟练啊,,)
1.
#include <bits/stdc++.h>using namespace std;const int N = 255;int a[55];long long dp[55][N];int main(){ int m,n; scanf("%d%d",&n,&m); for(int i = 1;i <= m;i++) scanf("%d",&a[i]); for(int i = 1;i <= m;i++) dp[i][0] = 1; for(int i = 1;i <= m;i++) for(int j = 1;j <= n;j++) if(a[i] > j) dp[i][j] = dp[i-1][j]; else dp[i][j] = dp[i-1][j] + dp[i][j-a[i]]; printf("%lld\n",dp[m][n]);}//dp[i][j]表示i种硬币组成和为j的方案数
2.
#include <bits/stdc++.h>using namespace std;const int N = 255;int a[55];long long dp[N];int main(){ int m,n; scanf("%d%d",&n,&m); for(int i = 1;i <= m;i++) scanf("%d",&a[i]); dp[0] = 1; for(int i = 1;i <= m;i++) for(int j = a[i];j <= n;j++) dp[j] += dp[j-a[i]]; printf("%lld\n",dp[n]);}//dp[i][j]表示i种硬币组成和为j的方案数
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