HDU 1231最大连续子序列（周赛F题）

这题挺气人的，比赛的时候编了提交了好多次都不知道哪里错了，搞了两天才知道自己的代码哪里错了看了队友的发现他的方法确实很好，参考他的方法……我试了用了下DP，但是就是提交不过，真的晕死……下面给出时间复杂度最低和第二低的算法，第一种为累计求和，时间复杂度为o(n)

#include <iostream>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        int a[10005],i,j=0,max=-1,sum=0,end,start=0,s,e;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(i=0;i<n;i++)
        {
            if(a[i]>=0) j=1;
            sum+=a[i];
            end=i;
            if(sum<0)
            {
                sum=0;
                start=i+1;
                end=i+1;
            }
            if(max<sum)
            {
                max=sum;
                s=start;
                e=end;
            }
        }
        if(j) printf("%d %d %d\n",max,a[s],a[e]);
        else printf("0 %d %d\n",a[0],a[n-1]);
    }
    return 0;
}

第二种为分治法，时间复杂度为0（n*logn)

#include <iostream>#include <map>#include <deque>#include <queue>#include <stack>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <map>#include <set>using namespace std;const int MAXN = 10010;struct ANS{    int sum;    int l, r;    ANS() {}    ANS(int a, int b, int c) : sum(a), l(b), r(c) {}};ANS maxsum(int* a, int x, int y){    int m, v, L, R, l, r;    ANS M(a[x], a[x], a[x]), t1, t2;    if (y - x == 1)        return M;    m = x + (y - x) / 2;    t1 = maxsum(a, x, m);    t2 = maxsum(a, m, y);    if (t1.sum >= t2.sum) M = t1;    else M = t2;    v = 0;    L = a[m - 1];    l = a[m - 1];    for (int i = m - 1; i >= x; i--)    {        v += a[i];        if (v > L) L = v, l = a[i];    }    v = 0;    R = a[m];    r = a[m];    for (int i = m; i < y; i++)    {        v += a[i];        if (v > R) R = v, r = a[i];    }    if (M.sum > (L + R))        return M;    return ANS(L + R, l, r);}int k, a[MAXN];int main(){    while (scanf("%d", &k) && k)    {        for(int i=0;i<k;i++)            scanf("%d", &a[i]);        ANS ans = maxsum(a, 0, k);        if (ans.sum >= 0)            printf("%d %d %d\n", ans.sum, ans.l, ans.r);        else            printf("0 %d %d\n", a[0], a[k - 1]);    }    return 0;}