最大子段和
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
题目大意：给出一串序列，要求输出其子序列的最大连续和与它所在的位置。
分析：由于时间限制为一秒且n=1e5，所用两个for循环肯定超时，不过可以动态规划，也可以二分。
代码如下：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int i,j,kase=0,n,t,b,c,d,x,sum;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(j=1;j<=n;j++)
{
scanf("%d",&x);
if(j==1)
{
sum=b=x;
i=c=d=1;
}
else
{
if(x>x+b)
{
b=x;
i=j;
}
else
b+=x;
}
if(b>sum)
{
sum=b;
c=i;
d=j;
}
}
printf("Case %d:
",++kase);
printf("%d %d %d
",sum,c,d);
if(t)
cout<<endl;
}
return 0;
}