题目：
Max Sum |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
Total Submission(s): 1390 Accepted Submission(s): 542 |
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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.
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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).
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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.
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Sample Input 25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5
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Sample Output Case 1:14 1 4Case 2:7 1 6
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Author Ignatius.L
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题目分析：
简单DP。这道题是求最大子序列的和，并且要求最大子序列的起点和终点。如果不需要求起点和终点的话。直接可以有last = max(0,last) + a[i]; ans = max(last,ans)来求解即可。但，其实就算要求起点和终点，思路还是一样的这是只不过不能就直接max(0,last)。应该把这个逻辑展开，求产生当前最大序列和的起点和终点。
代码如下：
/*
* a1.cpp
*
* Created on: 2015年2月6日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
using namespace std;
int main(){
int t;
scanf("%d",&t);
int k;
for(k = 1 ; k <= t ; ++k){
int n;
scanf("%d",&n);
int start = 1;//最大子序列和的起点.默认为1
int end = 1 ;//最大子序列和的终点.默认为1.
int thisSum = 0;//当前子序列的和
int maxSum = -100005;//最大子序列的和
int index = 1;//当前子序列的起点
int i;
int temp;
for(i = 1 ; i <= n ; ++i){
scanf("%d",&temp);
thisSum += temp;//当前子序列的和不断地增加
if(thisSum > maxSum){//如果当前子序列的和>最大子序列和
maxSum = thisSum;//更新最大子序列和
start = index;//将最大子序列的起点更新为当前子序列的起点
end = i;//将最大子序列的终点更新为当前遍历到的这个数(因为这个数产生了目前最大的子序列和)
}//需要注意的是,就算thisSum并没有比maxSum要大.这时候序列并没有断,要看一下后面是否会回升
/**
* 需要注意的是,前面不要把这两个if语句写成
* if(){
* }else if(){
* }的结构
*
* 因为thisSum < maxSum && thisSum < 0 的情况是存在的
*/
if(thisSum < 0){//如果后面没有回升,相反已经<0了
thisSum = 0;//断开原来的子序列,重新记数
index = i+1;//更新当前子序列的起点
}
}
printf("Case %d:\n",k);
printf("%d %d %d\n",maxSum,start,end);
if(k != t){
printf("\n");
}
}
return 0;
}
