Values whose Sum is 0

專題二(分治)131:19:34725:00:00

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A B C D E F G H ID - 4 Values whose Sum is 0

Time Limit:15000MS     Memory Limit:228000KB     64bit IO Format:%I64d & %I64u

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Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 228 ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6-45 22 42 -16-41 -27 56 30-36 53 -37 77-36 30 -75 -4626 -38 -10 62-32 -54 -6 45

Sample Output

5

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

#include <iostream>#include <numeric>#include <algorithm>#include <cstdio>using namespace std;int main(){    int n, m;    int day[100050];    scanf("%d %d", &n, &m);    int max_money = 0, min_money = 0;    for(int i = 0; i < n; ++i)    {        scanf("%d", &day[i]);        max_money +=day[i];        if(day[i] > min_money)        {            min_money = day[i];        }    }    int mid = max_money + min_money;    while(min_money < max_money)    {        mid = (min_money + max_money) / 2;        int k = 0;        int sum = 0;        for (int i = 0; i < n; i++)        {            sum += day[i];            if (sum > mid)            {                sum = day[i];                k++;            }        }        if (k < m) max_money = mid;        else min_money = mid + 1;    }    printf("%d",max_money);    return 0;}

還是二分，最大值是每天的和，最小值就是錢最多的那天，然後通過二分，看M是否符合要求，就這樣，繼續下一道了