Max Sum
Time Limit: 2000ms Memory limit: 32768K 有疑问？点这里^_^
题目描述
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.
输入
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).
输出
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.
示例输入
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
示例输出
Case 1: 14 1 4 Case 2: 7 1 6
提示
hdoj1003 注意：本题后台测试数据量比较大，请使用C语言scanf输入。避免超时！！！！
来源
Ignatius.L
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
using namespace std;
int a[100100],b[100100];
int main()
{
int T;
int k = 0;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
b[0] = a[0];
for(int i=1;i<n;i++)
{
if(b[i-1]<0)
{
b[i] = a[i];
}
else
{
b[i] = a[i] + b[i-1];
}
}
int xx,yy;
int maxx = -999999;
for(int i=0;i<n;i++)
{
if(maxx<b[i])
{
maxx = b[i];
yy = i+1;
}
}
int pmax = 0;
for(int i=yy-1;i>=0;i--)
{
pmax += a[i];
//printf("pmax = %d maxx = %d\n",pmax,maxx);
if(pmax == maxx)
{
xx = i+1;
// break;
}
}
printf("Case %d:\n",++k);
printf("%d %d %d\n",maxx,xx,yy);
if(T)
{
printf("\n");
}
}
return 0;
}
