hdu1003 Max Sum (求连续的和最大的自链

Max SumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 242181    Accepted Submission(s): 57180Problem DescriptionGiven 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. InputThe 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). OutputFor 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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5 Sample Output

Case 1:14 1 4Case 2:7 1 6题意：t组数据，长度为n的字符串，找出一段连续且和最大的字符串#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;const int inf=0x3f3f3f3f;int main(){int t;scanf("%d",&t);int k=t;while(t--){int n;scanf("%d",&n);int maxx=-inf,sum=0;int x=1,y=0,temp=1;//temp用来标记起点for(int i=0; i<n; i++){int a;scanf("%d",&a);sum+=a;if(sum>maxx)//两个if判断不能换位置，例5 -1 -2 -3 -4 -5//这样会提前改变sum和temp的值，从而错误{maxx=sum;x=temp;y=i+1;}if(sum<0){sum=0;temp=i+2;//如果不用temp标记起点，直接用x的话例：5 -1 -2 -3 -4 -5//x会一直加下去，输出就会出错//x=i+1;}}printf("Case %d:\n%d %d %d\n",k-t,maxx,x,y);if(t!=0)printf("\n");}return 0;}

这个是错误的代码，我考虑一下我错哪了#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;const int inf=0x3f3f3f3f;int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int maxx=-inf,sum=0;int x=0,y=0,temp=1;//这种代码，有特别多的弊端，//值全为0时，maxbc0dx，x，y都不对//还有只要sum>0都能进入那个循环，y就会一直加,也是错的for(int i=0;i<n;i++){int a;scanf("%d",&a);sum+=a;if(sum<0){x=i+1;y=i+1;sum=0;}else{maxx=max(maxx,sum);y+=1;}}}}