HDU 1003 Max Sum(最大子段和+路径)

xiaoxiao2021-02-28  17

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 281619    Accepted Submission(s): 66887 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

题意:输出最大子段和以及元素的起始、终止位置

#include<stdio.h> #include<algorithm> using namespace std; int a[100002]; int n; void maxsum() { int s=0;//起点 int e=0;//终点 int sum=0; int maxn=-99999; int k=0; for(int i=0;i<n;i++) { sum+=a[i]; if(maxn<sum) { maxn=sum; s=k; e=i; } if(sum<0) { sum=0; k=i+1; } } //这里s+1,e+1,是因为此题下标以一开始 printf("%d %d %d\n",maxn,s+1,e+1); } int main() { int t; scanf("%d",&t); int ca=1; while(t--) { scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&a[i]); } printf("Case %d:\n",ca++); maxsum(); if(t) printf("\n"); } }
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