Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 252943 Accepted Submission(s): 60001
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
题目的意思是求一段连续的和最大序列，如果有多个序列，则输出第一个。
代码思想用当前部分和与之前的最大部分和比较，若比之前的大，则更新。
若当前部分和小于零，即可以废弃当前部分和，寻求下一个。
最后判断全是负数的情况
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 using namespace std; 5 6 int main() 7 { 8 //freopen("in.txt","r",stdin); 9 int T, n, subsum, ans; //subsum：当前部分和，ans：答案 10 int N = 1; 11 int Sta, End, Sta1; //Sta：至今最大和起始位置，End：至今最大和终止位置，Sta1当前部分和的起始位置 12 int a[110000]; 13 scanf("%d",&T); 14 while( T--) { 15 scanf("%d",&n); 16 if(N != 1) printf(" "); 17 printf("Case %d: ",N++); 18 ans = -1, Sta = End = Sta1 = 1, subsum = 0; 19 for(int i = 1 ; i <= n ; i++ ) { 20 scanf("%d",&a[i]); 21 subsum += a[i]; 22 if(subsum < 0) {Sta1 = i+1 ; subsum = 0;}//当前部分和是负数，则遗弃当前段部分和，寻找下一段 23 else { 24 if(subsum > ans) {//当前部分和大于之前的某一段部分和，则更新 25 Sta = Sta1; 26 End = i; 27 ans = subsum ; 28 } 29 } 30 31 } 32 if(ans == -1) {//判断全是负数的情况 33 ans = -100000; 34 for(int i = 1; i <= n; i++) { 35 if(ans < a[i]) { 36 ans = a[i]; 37 Sta = End = i; 38 } 39 } 40 printf("%d %d %d ",ans ,Sta, End); 41 } 42 else 43 printf("%d %d %d ",ans, Sta, End); 44 } 45 return 0; 46 }