C. George and Job time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.
Given a sequence of n integers p1,?p2,?...,?pn. You are to choose k pairs of integers:
[
l
1,?
r
1],?[
l
2,?
r
2],?...,?[
l
k,?
r
k] (1?≤?
l
1?≤?
r
1?
l
2?≤?
r
2?
l
k?≤?
r
k?≤?
n;
r
i?-?
l
i?+?1?=?
m),?
in such a way that the value of sum 
is maximal possible. Help GeZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcmdlIHRvIGNvcGUgd2l0aCB0aGUgdGFzay48L3A+CgoKSW5wdXQKPHA+VGhlIGZpcnN0IGxpbmUgY29udGFpbnMgdGhyZWUgaW50ZWdlcnMgPGVtPm48L2VtPiwKPGVtPm08L2VtPiBhbmQgPGVtPms8L2VtPgooMT+h3D8oPGVtPm08L2VtPj+hwT88ZW0+azwvZW0+KT+h3D88ZW0+bjwvZW0+P6HcPzUwMDApLiBUaGUgc2Vjb25kIGxpbmUgY29udGFpbnMKPGVtPm48L2VtPiBpbnRlZ2VycyA8ZW0+cDwvZW0+PHN1YiBjbGFzcz0="lower-index">1,?p2,?...,?pn (0?≤?pi?≤?109).
Output
Print an integer in a single line ― the maximum possible value of sum.
Sample test(s) Input
5 2 1
1 2 3 4 5
Output
9
Input
7 1 3
2 10 7 18 5 33 0
Output
61
情景就不赘述了，给你n个数p1~pn，要你求出这个数列中k段长为m的区间的元素之和的最大值。
要求区间元素和，当然想求出数列的前缀和来了。
二维dp，dp[i][j]表示在取到第i个元素的时候取j个m长度的区间之和的最大值。
dp[i][j]可由dp[i-1][j]不添加区间得到的,或者由dp[i-m][j-1]添加一个区间得到。
遂有状态转移方程：dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+s[i]-s[i-m])，其中s是前缀和。
然后就是简单题了。
#include
#include
#include
#include
using namespace std; long long dp[5005][5005]; int main(){ int n,m,k; scanf ("%d%d%d",&n,&m,&k);// long long a[5005],s[5005]; memset(a,0,sizeof(a)); memset(s,0,sizeof(s)); memset(dp,0,sizeof(dp)); for (int i=1;i<=n;i++){ scanf ("%I64d",&a[i]); s[i]=s[i-1]+a[i]; } for (int i=m;i<=n;i++){ for (int j=1;j<=k;j++){ dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+s[i]-s[i-m]); } } printf ("%I64d\n",dp[n][k]); return 0; }
（ps：数据爆int） :(