HDU 1024 Max Sum Plus Plus(基礎dp)
阿新 • • 發佈:2018-05-01
always cut panel face make return algo pan nbsp
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2 ) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don‘t want to write a special-judge module, so you don‘t have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^ , S2, S3 ... Sn.
Process to the end of file.
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34541 Accepted Submission(s): 12341
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2
But I`m lazy, I don‘t want to write a special-judge module, so you don‘t have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Input Each test case will begin with two integers m and n, followed by n integers S1
Process to the end of file.
Output Output 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.
Author JGShining(極光炫影)
Recommend We have carefully selected several similar problems for you: 1074 1025 1081 1080 1160 原題地址 題意:給我們一個長度為N的數組讓我們把數組分成M個不想交的字串 使得M個字串的和最大
#include <iostream> #include<cstdio> #include<algorithm> using namespace std; #define MAXN 1100000 #define INF 0x3f3f3f3f int dp[MAXN]; int maxn[MAXN]; int a[MAXN]; int main() { int n,m; std::ios::sync_with_stdio(false); while(cin>>m>>n){ for(int i=1;i<=n;i++){ cin>>a[i]; maxn[i]=0; dp[i]=0; } dp[0]=0; maxn[0]=0; int maxx; for(int i=1;i<=m;i++){ maxx=-INF; for(int j=i;j<=n;j++){ dp[j]=max(dp[j-1]+a[j],maxn[j-1]+a[j]); maxn[j-1]=maxx; maxx=max(maxx,dp[j]); } } cout<<maxx<<endl; } return 0; }
HDU 1024 Max Sum Plus Plus(基礎dp)