hdu 1024 Max Sum Plus Plus DP
Max Sum Plus PlusTime Limit: 1 Sec Memory Limit: 256 MB题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=1024
Description
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. ^_^
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.
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
题意
给你n个数，让你选择连续的M段，让你得到最大值
题解：
简单的想一想，dp表示，前j个数，我选i段的最大值，那么转移方程,dp=max(dp+a,dp+a)
然后滚动数组优化一下
代码：
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1000005
#define mod 10007
#define eps 1e-9
int Num;
char CH;
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/*
inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH+48);
puts("");
}
*/
//**************************************************************************************
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH+48);
puts("");
}
int dp;
int mmax;
int a;
int main()
{
int n,m;
int i,j,mmmax;
while(scanf("%d%d",&m,&n)!=EOF)
{
for(i=1;i<=n;i++)
{
scanf("%d",&a);
mmax=0;
dp=0;
}
dp=0;
mmax=0;
for(i=1;i<=m;i++)
{
mmmax=-inf;
for(j=i;j<=n;j++)
{
dp=max(dp+a,mmax+a);
mmax=mmmax;
mmmax=max(mmmax,dp);
}
}
printf("%d\n",mmmax);
}
return 0;
}
文档来源：51CTO技术博客https://blog.51cto.com/u_15303184/3111735
页:
[1]