HDU 1003 Max Sum 最大连续子序列和

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Given a sequence a[1],a[2],a[3]……a
, 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

最大子序列和，输出第一个和最后一个元素的位置，有相同的需要输出第一次出现的位置和最后一次出现的位置。

既然找最大子序列和，那么最大子序列最后一个必然是存在这个序列中的位置，不妨把每一个数字假设成最大子序列的最后一个位置，dp[1] = a[1]，那么从dp[2]开始，dp[2]是最后一个，和它前面的dp[1]比较，如果dp[1]是正数，则dp[2] = a[2]+dp[1]，如果是负数，就不加，dp[2] = a[2]。如果dp[1]是负数，a[2]+dp[1] < a[2]，所以可以写成dp[2] = max(dp[1]+a[2],a[2]); 对后面么一个元素都这么推，就可以推出状态转移方程dp[i] = max(dp[i-1]+a[i],a[i]);

#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"string.h"
#include"stdlib.h"
#include"queue"
#define inf 7777777
using namespace std;

int a[100010];
int dp[100010];
int main(void)
{
int T;
scanf("%d",&T);
for(int j = 1;j <= T;j++)
{
int n;
scanf("%d",&n);
scanf("%d",&a[1]);
int maf;   ///最大值
maf = dp[1] = a[1];
int loc2 = 1;
int loc1;
for(int i = 2;i <= n;i++)
{
scanf("%d",&a[i]);
dp[i] = max(dp[i-1]+a[i],a[i]);
if(maf <= dp[i])     ///记录下最大dp值和出现位置
{
maf = dp[i];
loc2 = i;
}
}
for(int i = loc2,t = maf;i > 0;i--)
{
t -= a[i];
if(t == 0)   ///一直循环，每一次 t==0 的时候就是一个位置，最后一次的时候就是第一次出现的位置
{
loc1 = i;
}
}
printf("Case %d:\n%d %d %d\n",j,maf,loc1,loc2);
if(j != T)
printf("\n");    ///最后一组数据不能输出/n
}
return 0;
}