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 = 0, ch;

    while (!((ch = getchar()) >= '0' && ch <= '9')) {

        if (ch == EOF)

            return -1;

    }

    res = ch - '0';

    while ((ch = getchar()) >= '0' && ch <= '9')

        res = res * 10 + (ch - '0');

    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 = 0, ch, flag = 0;

    while (!((ch = getchar()) >= '0' && ch <= '9')) {

        if (ch == '-')

            flag = 1;

        if (ch == EOF)

            return -1;

    }

    res = ch - '0';

    while ((ch = getchar()) >= '0' && ch <= '9')

        res = res * 10 + (ch - '0');

    if (flag == 1)

        res = -res;

    return res;

}



const int MAXN = 10009;

int maxr[MAXN], minr[MAXN];



inline int mymul(int a, int b) {

    int sign = 1;

    if (a * b == 0) {

        return 0;

    }

    if (a * b < 0) {

        sign = -1;

    }

    return sign * (abs(a) + abs(b));

}



int main() {

    int T = get_int(), tmp, N;

    int a[3];

    for (int t = 1; t <= T; t++) {

        N = get_int();

        maxr[0] = minr[0] = 0;

        int ans = 0;

        for (int i = 1; i <= N; i++) {

            tmp = get_int2() / 2;

            a[0] = mymul(maxr[i - 1], tmp);

            a[1] = mymul(minr[i - 1], tmp);

            a[2] = tmp;

            sort(a, a + 3);

            maxr[i] = a[2];

            ans = maxr[i] > ans ? maxr[i] : ans;

            minr[i] = a[0];

        }

        printf("Case #%d: %d\n", t, ans);

    }

    return 0;

}