标签：fir ali careful osi amp which 最大的 should ref
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
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如果都是负数，自然找到最大的负数即可，否则可以用动态规划法，如果sum<0了，那就没必要继续存了，只会影响了后面的，把后面的变小了，此时置sum为0.
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define inf 0x3f3f3f3f
#define MAX 1000
using namespace std;
int n,m,d;
int main() {
scanf("%d",&n);
for(int i = 1;i <= n;i ++) {
if(i > 1) putchar(‘\n‘);
scanf("%d",&m);
int l = 1,sum = 0,maxsum = 0,a = 0,b = 0,maxnum = -inf,t;
for(int j = 1;j <= m;j ++) {
scanf("%d",&d);
if(d > maxnum) {
maxnum = d;
t = j;
}
sum += d;
if(sum < 0) {
sum = 0;
l = j + 1;
continue;
}
if(sum > maxsum) {
maxsum = sum;
a = l;
b = j;
}
}
if(!a) {
a = b = t;
maxsum = maxnum;
}
printf("Case %d:\n%d %d %d\n",i,maxsum,a,b);
}
}
hdu 1003 Max Sum
标签：fir ali careful osi amp which 最大的 should ref
原文地址：https://www.cnblogs.com/8023spz/p/9748903.html
