标签：always include think test follow and integer 动态 more
InputEach test case will begin with two integers m and n, followed by n integers S 1, S2, S 3 ... S n.
Process to the end of file.
OutputOutput the maximal summation described above in one line.
Sample Input
1 3 1 2 3 2 6 -1 4 -2 3 -2 3
Sample Output
6
8
Hint
Huge input, scanf and dynamic programming is recommended.
若不做任何优化，并不考虑数据大小，仅考虑样例
#include<iostream> #include<algorithm> using namespace std; int num[1000]; int dp[10][100]; int main(){ int n,m; while(cin>>n>>m){ for(int i=1;i<=m;i++){ cin>>num[i]; } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ int m=-1; for(int w=1;w<j;w++){ m=max(dp[i-1][w],m); } dp[i][j]=max(m,dp[i][j-1])+num[j]; } } int ans=-1; for(int i=1;i<=m;i++){ ans=max(ans,dp[n][i]); } cout<<ans<<endl; } return 0; }
#include<iostream> #include<algorithm> #include<string.h> using namespace std; int dp[1000005]; int Max[1000005]; int num[1000005]; int main(){ int m,n; while(cin>>m>>n){ int k; for(int i=1;i<=n;i++){ cin>>num[i]; } memset(dp,0,sizeof(dp)); memset(Max,0,sizeof(Max)); int mmax; for(int i=1;i<=m;i++){ mmax=-INT_MAX; for(int j=i;j<=n;j++){ dp[j]=max(dp[j-1],Max[j-1])+num[j]; Max[j-1]=mmax; mmax=max(mmax,dp[j]); } } cout<<mmax<<endl; } return 0; }
HDOJ1024--Max Sum Plus Plus(动态规划)UnSolved
标签：always include think test follow and integer 动态 more
原文地址：https://www.cnblogs.com/albert67/p/10332175.html
