二分-G - 4 Values whose Sum is 0
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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 2 28 ) 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 45Sample Output
5Hint
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).
题目大意:每行会给出四个整数,要求在每列整数中找出一个整数,使四个整数之和为0,求有几种不同组合
1 #include<iostream> 2 #include<algorithm> 3 using namespace std; 4 5 const int maxn = 4010; 6 const int maxm = 4010*4010; 7 int a[maxn],b[maxn],c[maxn],d[maxn]; 8 int ab[maxm],cd[maxm]; 9 10 int main(){ 11 int n; 12 scanf("%d",&n); 13 for(int i=0;i<n;i++) 14 scanf("%d %d %d %d",a+i,b+i,c+i,d+i); 15 16 int cnt = 0; 17 for(int i=0;i<n;i++)//前两列整数求和 18 for(int j=0;j<n;j++) 19 ab[cnt++] = a[i] + b[j]; 20 cnt = 0; 21 for(int i=0;i<n;i++)//后两列整数求和 22 for(int j=0;j<n;j++) 23 cd[cnt++] = c[i] + d[j]; 24 25 sort(ab,ab+n*n); 26 sort(cd,cd+n*n); 27 28 int p = n*n-1; 29 cnt = 0; 30 for(int i=0;i<n*n;i++){ 31 for(;p>=0;p--) 32 if(ab[i]+cd[p]<=0) break; 33 if(p<0) break; 34 for(int j=p;j>=0;j--){ 35 if(ab[i]+cd[j]==0) cnt++; 36 if(ab[i]+cd[j]<0) break; 37 } 38 } 39 printf("%d",cnt); 40 }
面对四列整数,一一组合将会造成极其庞大的数据,可将其中两两组合,组合以后再相加寻找和为0的情况。
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