HDU 1003 - Max Sum

从O(n3)的最普通的方法,用数学数列方法优化到O(n2),最后优化到O(n),终于AC了,最后的这个确实不好想。

O(n3):

#include <iostream>
#include <cstdio>
using namespace std;

int a[102400];

int main()
{
    ios::sync_with_stdio(false);
    int T, kase = 0; cin >> T;
    while(T--){
        int N; cin >> N;
        for(int i = 1; i <= N; ++i)
            cin >> a[i];
        int best = a[0], l = 1, r = 1;
        for(int i = 1; i <= N; ++i)
            for(int j = i; j <= N; ++j){
                int sum = 0;
                for(int k = i; k <= j; ++k)
                    sum += a[k];
                if(sum > best){
                    best = sum;
                    l = i; r = j;
                }
            }
        if(kase) printf("\n");
        printf("Case %d:\n", ++kase);
        printf("%d %d %d\n", best, l, r);
    }
    return 0;
}

O(n2):

#include <iostream>
#include <cstdio>
using namespace std;

int a[102400];
int s[102400];

int main()
{
    ios::sync_with_stdio(false);
    int T, kase = 0; cin >> T;
    while(T--){
        int N; cin >> N;
        s[0] = 0;
        for(int i = 1; i <= N; ++i){
            cin >> a[i];
            s[i] = s[i-1] + a[i];
        }
        int best = a[1], l = 1, r = 1;
        for(int i = 1; i <= N; ++i)
            for(int j = i; j <= N; ++j){
                int sum = s[j] - s[i-1];
                if(sum > best){
                    best = sum;
                    l = i; r = j;
                }
            }
        if(kase) printf("\n");
        printf("Case %d:\n", ++kase);
        printf("%d %d %d\n", best, l, r);
    }
    return 0;
}

O(n):

#include <iostream>
#include <cstdio>
using namespace std;

int a[102400];

int main()
{
    ios::sync_with_stdio(false);
    int T, kase = 0; cin >> T;
    while(T--){
        int N, sum = 0, max = INT_MIN, l = 1, r = 1, t = 1; cin >> N;
        for(int i = 1; i <= N; ++i){
            cin >> a[i]; sum += a[i];
            if(sum > max) max = sum, l = t, r = i;
            if(sum < 0) t = i + 1, sum = 0;
        }
        if(kase) printf("\n");
        printf("Case %d:\n", ++kase);
        printf("%d %d %d\n", max, l, r);
    }
    return 0;
}


posted @ 2015-02-19 20:21  Popco  阅读(101)  评论(0编辑  收藏  举报