hdu4561 bjfu1270 最大子段积

就是最大子段和的变体。最大子段和只要一个数组，记录前i个里的最大子段和在f[i]里就行了，但是最大子段积因为有负乘负得正这一点，所以还需要把前i个里的最小子段积存起来。就可以了。直接上代码：

/*
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
//输入非负整数，用法int a = get_int();
int get_int() {
    int res = , ch;
    while (!((ch = getchar()) >= '' && ch <= '')) {
        if (ch == EOF)
            return -;
    }
    res = ch - '';
    while ((ch = getchar()) >= '' && ch <= '')
        res = res *  + (ch - '');
    return res;
}
//输入整数(包括负整数，故不能通过返回值判断是否输入到EOF，本函数当输入到EOF时，返回-1)，用法int a = get_int2();
int get_int2() {
    int res = , ch, flag = ;
    while (!((ch = getchar()) >= '' && ch <= '')) {
        if (ch == '-')
            flag = ;
        if (ch == EOF)
            return -;
    }
    res = ch - '';
    while ((ch = getchar()) >= '' && ch <= '')
        res = res *  + (ch - '');
    if (flag == )
        res = -res;
    return res;
}

const int MAXN = ;
int maxr[MAXN], minr[MAXN];

inline int mymul(int a, int b) {
    int sign = ;
    if (a * b == ) {
        return ;
    }
    if (a * b < ) {
        sign = -;
    }
    return sign * (abs(a) + abs(b));
}

int main() {
    int T = get_int(), tmp, N;
    int a[];
    for (int t = ; t <= T; t++) {
        N = get_int();
        maxr[] = minr[] = ;
        int ans = ;
        for (int i = ; i <= N; i++) {
            tmp = get_int2() / ;
            a[] = mymul(maxr[i - ], tmp);
            a[] = mymul(minr[i - ], tmp);
            a[] = tmp;
            sort(a, a + );
            maxr[i] = a[];
            ans = maxr[i] > ans ? maxr[i] : ans;
            minr[i] = a[];
        }
        printf("Case #%d: %d\n", t, ans);
    }
    return ;
}