4 Values whose Sum is 0 POJ - 2785
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2022-05-25 16:25:26
4 Values whose Sum is 0 Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 29243 Accepted: 8887 Case Time Limit: 5000MS Description The SUM ......
4 values whose sum is 0
time limit: 15000ms | memory limit: 228000k | |
total submissions: 29243 | accepted: 8887 | |
case time limit: 5000ms |
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 -46 26 -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).
题意:abcd四个数列,每个数列抽一个数字,使得和为0,问一共有几组。当一个数列中有多个相同的数字时,把他们作为不同的数字看待。
思路:直接爆搜肯定会超时,将它们对半分成ab和cd再考虑就可以解决。比如在ab中取出a,b后,cd中要取出-a-b,因此先把cd中所有取数字的情况n*n种给纪录下来,然后排序,用二分查找次数。
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<cstdlib> #include<queue> #include<set> #include<vector> using namespace std; #define inf 0x3f3f3f3f #define eps 1e-10 typedef long long ll; const int maxn = 4002; const int mod = 1e9 + 7; int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } int n; int a[maxn],b[maxn],c[maxn],d[maxn]; int cd[maxn*maxn]; void solve() { for(int i=0;i<n;i++) for(int j=0;j<n;j++) { cd[i*n+j]=c[i]+d[j]; } sort(cd,cd+n*n); ll res=0; for(int i=0;i<n;i++) for(int j=0;j<n;j++) { int ab=-(a[i]+b[j]); res+=upper_bound(cd,cd+n*n,ab)-lower_bound(cd,cd+n*n,ab); } cout<<res<<endl; } int main() { scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]); solve(); }
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