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二分基础
阅读量:7044 次
发布时间:2019-06-28

本文共 1613 字,大约阅读时间需要 5 分钟。

/*A - 二分法 基础Time Limit:15000MS     Memory Limit:228000KB     64bit IO Format:%I64d & %I64uSubmit StatusDescriptionThe 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 .InputThe 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 2 28 ) that belong respectively to A, B, C and D .OutputFor each input file, your program has to write the number quadruplets whose sum is zero.Sample Input6-45 22 42 -16-41 -27 56 30-36 53 -37 77-36 30 -75 -4626 -38 -10 62-32 -54 -6 45Sample Output5HintSample 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).By Grant Yuan2014.7.14二分*/#include
#include
#include
#include
#include
#include
using namespace std;long long a[4002],b[4002],c[4002],d[4002];int n;long long num1[16000004];long long num2[16000004];int top;long long sum;int binary(long k,int left,int right){ int i; while(left<=right){ int mid=(left+right)/2; int num=0; if(num2[mid]==k) { num=1; for(i=mid-1;i>=0&&num2[i]==k;i--) num++; for(i=mid+1;i
k) right=mid-1; else left=mid+1; } return 0;}int main(){ cin>>n; long long flag; for(int i=0;i
>a[i]>>b[i]>>c[i]>>d[i]; top=-1; sum=0; for(int i=0;i

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