最大子序列和

O(n)：DP O(nlgn)：分治法 O(n²)：列舉

HDU 1003

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

AC-O(n)

#include <bits/stdc++.h>
using namespace std;
int main(){
    int a[100010],T,N,position,beginum,endnum,maxsum,thissum;
    cin >>T;
    for(int i=0;i<T;i++){
        cin >>N;
        for(int j=0;j<N;j++)
            cin >>a[j];
        position = beginum = endnum =0;
        maxsum = thissum = a[0];
		for(int k=1;k<N;k++){
			if(thissum + a[k] < a[k]){             //如果當前值比a[i]小的話則改為a[i]
				thissum = a[k];
				position = k;                  //記錄下改的位置
			}
			else{
				thissum = thissum + a[k];
			}
			if(thissum > maxsum){                //當前值比最大值大，則頭尾都要改
				maxsum = thissum;
				beginum = position;
				endnum = k;
			}
		}
		if(i)
            cout <<endl;
		cout <<"Case "<<i+1<<':'<<endl;
		cout <<maxsum<<' '<<beginum+1<<' '<<endnum+1<<endl;
    }
    return 0;
}
 

AC-O(nlgn)

//O(nlogn)   分治排序演算法
int maxsum_version_1(int *a,int x,int y){  //求區間[x,y)的最大值     [2,3)為a2;
	int L,R,v,m,i,max;
	if(y-x==1)
		return a[x];
	m=x+(y-x)/2;                 //不取m=(y-x)/2 是為了使分界點靠近區間左端點
	max=(maxsum_version_1(a,x,m)>maxsum_version_1(a,m,y))?maxsum_version_1(a,x,m):maxsum_version_1(a,m,y);
	v=0;
	L=a[m-1];
	for(i=m-1;i>=x;i--){
		v=v+a[i];
		L=L>v?L:v;
	}
	R=a[m];
	v=0;
	for(i=m;i<y;i++){
		v+=a[i];
		R=R>v?R:v;
	}
	return (max>L+R)?max:L+R;