Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 161294 Accepted Submission(s): 37775
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
Author
Ignatius.L
Recommend
题意：给出n个数的序列，求出最大的子串和。并输出起点和终点。
思路：dp[i]表示以i为结尾的最大子串和。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 100005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
int dp[maxn]; //dp[i]表示以i为结尾的最大子串和
int a[maxn];
int n;
int main()
{
int i,j,t,cas=1;;
sf(t);
bool flag=false;
while (t--)
{
if (flag) pf("\n");
flag=true;
sf(n);
FRL(i,1,n+1)
sf(a[i]);
mem(dp,0);
dp[1]=a[1];
int S=1,T=1,s=1,t=1,maxx=a[1];//s,t记录当前首尾指针。S。T记录当前最大值的首尾指针
FRL(i,2,n+1)
{
if (a[i]>dp[i-1]+a[i]) //假设a[i]对dp[i-1]没有贡献反而会使dp[i-1]减小，那么就以i又一次作为起点
{
s=i;
t=i;
dp[i]=a[i];
}
else if (a[i]<=dp[i-1]+a[i]) //a[i]比dp[i-1]+a[i]更大，就将尾指针t向后移赋为i
{
t=i;
dp[i]=dp[i-1]+a[i]; //更dp[i]
}
if (dp[i]>maxx) //假设当前dp[i]比之前的最大值要大，更新最大值，并记录首尾指针
{
S=s;
T=t;
maxx=dp[i];
}
}
pf("Case %d:\n",cas++);
pf("%d %d %d\n",maxx,S,T);
}
return 0;
}
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5077522.html，如需转载请自行联系原作者
