题目链接:
https://cn.vjudge.net/problem/POJ-2785
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 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).
1 /* 2 问题 给出n行的4个数,这四列数分别是A,B,C,D的集合,问有多少组ABCD相加和为0 3 解题思路 刚开始没读懂题就开始写了,没想到题意是另一个意思,还是按练习要求做题吧。 4 读懂了题,脑子里马上跳出4重循环,又一看n最大为4000,还是放弃吧。 5 看了一下分析,先将a+b的结果与其出现的次数放在map容器里,再将c+d的结果与其出现的次数放在map容器里,最后查找一下, 6 如果存在则累计结果。但是超时,原因是常数较大时使用map也可能超时。 7 随后在网上看到一种更为巧妙的解法,将C和D的所有结果存放在一个一维数组中,再将其排序,遍历A+B的和,累加在这个二维数组 8 中的个数即可。 9 */ 10 11 /*解法一 超时!!! 12 #include<cstdio> 13 #include<iostream> 14 #include<map> 15 using namespace std; 16 17 int main(){ 18 int T,n,cou,i,j,a[4010],b[4010],c[4010],d[4010]; 19 map<int,int> m1,m2; 20 21 while(scanf("%d",&n) != EOF) 22 { 23 j=0; 24 for(i=1;i<=n;i++){ 25 scanf("%d%d%d%d",&a[j],&b[j],&c[j],&d[j]); 26 j++;//不能缩放在上面的一句 27 } 28 29 for(i=0;i<n;i++){ 30 for(j=0;j<n;j++){ 31 m1[ a[i]+b[j] ]++; 32 } 33 } 34 for(i=0;i<n;i++){ 35 for(j=0;j<n;j++){ 36 m2[ -1*(c[i]+d[j] ) ]++; 37 } 38 } 39 40 map<int,int>::iterator it1,it2; 41 int ans=0; 42 for(it1=m1.begin(); it1 != m1.end(); it1++){ 43 it2=m2.find(it1->first); 44 if(it2 != m2.end()){ 45 ans += (it1->second * it2->second); 46 } 47 } 48 printf("%d\n",ans); 49 } 50 return 0; 51 }*/ 52 //解法二 53 #include<cstdio> 54 #include<algorithm> 55 using namespace std; 56 57 int cd[4010*4010];//一维数组当二维数组用 58 59 int main(){ 60 int T,n,cou,i,j,a[4010],b[4010],c[4010],d[4010],sumab; 61 long long ans; 62 while(scanf("%d",&n) != EOF) 63 { 64 for(i=0;i<n;i++) 65 scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]); 66 67 for(i=0;i<n;i++){ 68 for(j=0;j<n;j++){ 69 cd[i*n+j]=c[i]+d[j]; 70 } 71 } 72 73 sort(cd,cd+n*n); 74 75 ans=0; 76 for(i=0;i<n;i++){ 77 for(j=0;j<n;j++){ 78 sumab=-1*(a[i]+b[j]); 79 ans += upper_bound(cd,cd+n*n,sumab) - lower_bound(cd,cd+n*n,sumab); 80 //使用参数,起点+终点+目标值 81 } 82 } 83 84 printf("%lld\n",ans); 85 } 86 return 0; 87 } 88
原文地址:https://www.cnblogs.com/wenzhixin/p/8733441.html