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
Author
Ignatius.L
题目大意:求最大字段和
n<=100000,t<=20,可知只能使用线性复杂度的算法。
设f[i]表示结尾为i的最大子段和,f[i]=max{f[i-1]+a[i],a[i]}
ans=max{f[i]}
#include<iostream> #include<cstring> #include<climits> #include<cstdio> using namespace std; const int N=100005; int T,n,t,a[N],f[N],res,x,y,pth[N]; int main() { scanf("%d",&T); while(T--) { printf("Case %d: ",++t); scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",a+i); res=f[0]=INT_MIN; for(int i=1;i<=n;i++) { if(f[i-1]<0) { f[i]=a[i]; pth[i]=i; } else { f[i]=a[i]+f[i-1]; pth[i]=pth[i-1]; } if(f[i]>res) { res=f[i]; x=pth[i],y=i; } } printf("%d %d %d ",res,x,y); if(T) printf(" "); } return 0; }