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 S
1, S
2, S
3, S
4 ... S
x, ... S
n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S
x ≤ 32767). We define a function sum(i, j) = S
i + ... + S
j (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(i
1, j
1) + sum(i
2, j
2) + sum(i
3, j
3) + ... + sum(i
m, j
m) maximal (i
x ≤ i
y ≤ j
x or i
x ≤ j
y ≤ j
x 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(i
x, j
x)(1 ≤ x ≤ m) instead. ^_^
Each test case will begin with two integers m and n, followed by n integers S
1, S
2, S
3 ... S
n.
Process to the end of file.
Output the maximal summation described above in one line.
Huge input, scanf and dynamic programming is recommended.
#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;
}
