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.<br>
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).<br>
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.<br>
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
解题思路:
给定一串数字,求其中某段连续的数字和最大的那一段,输出最大子段的和以及这个字段的起始和结束位置
用贪心算法,第一个数时最大的数是第一个数,第二个数时与第一个数想加,得到的数与第二个数比较,如果最大的是第二个数,从第二个数开始,反之继续相加
代码:
#include<iostream> #define N 100010 using namespace std; int a[N],d[N]; int main() { int test,n,i,max,k,f,e; cin>>test; k=1; while(test--) { cin>>n; for(i=1;i<=n;i++) cin>>a[i]; d[1]=a[1]; for(i=2;i<=n;i++) { if(d[i-1]<0) d[i]=a[i]; else d[i]=d[i-1]+a[i]; } max=d[1];e=1; for(i=2;i<=n;i++) { if(max<d[i]) { max=d[i];e=i; } } int t=0; f=e; for(i=e;i>0;i--) { t=t+a[i]; if(t==max) f=i; } cout<<"Case "<<k++<<":"<<endl<<max<<" "<<f<<" "<<e<<endl; if(test) cout<<endl; } return 0; }