Max Sum Plus Plus——A
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A. Max Sum Plus Plus
Now I think you have got an AC in Ignatius.L‘s "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.
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. ^_^
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. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
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 (子段1: 4;子段2:3 -2 3)
Sample Output
6 8
Hint
Huge input, scanf and dynamic programming is recommended.
题意:求最大M子段和
#include <iostream> #include<cmath> #include<cstring> using namespace std; const int MAX=1000010; const int INF=0x7fffffff; int a[MAX]; int b[MAX]; int c[MAX]; int main() { int m,n; while(cin>>n>>m) { for(int i=1;i<=m;i++) cin>>a[i]; memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); int maxn; for(int i=1;i<=n;i++) { maxn=(-1)*INF; for(int j=i;j<=m;j++) { b[j]=max(b[j-1]+a[j],c[j-1]+a[j]); c[j-1]=maxn; if(b[j]>maxn) maxn=b[j]; } } cout<<maxn<<endl; } return 0; }
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