【POJ - 2785】4 Values whose Sum is 0 (二分,折半枚举)

xiaoxiao2025-11-17  9

题干:

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).

解题报告:

   这题要求和为0,显然直接暴力是不存在的,但是还是有解题线索的,比如说一共就四个数,显然可以折半啊!!相同思路的遇到过一次了吧?题目名叫“暴力AC”,写过那篇博客但是具体哪场比赛忘记了。

AC代码:

#include<cstdio> #include<iostream> #include<algorithm> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll a[4005],b[4005],c[4005],d[4005]; int tot1=0,tot2=0; ll ans1[16000000],ans2[16000000]; int get(ll x) { return lower_bound(ans2+1,ans2+tot2+1,x) - ans2; } int main() { int n; while(cin>>n) { for(int i = 1; i<=n; i++) { scanf("%lld%lld%lld%lld",a+i,b+i,c+i,d+i); } tot1=tot2=0; for(int i = 1; i<=n; i++) { for(int j = 1; j<=n; j++) { ans1[++tot1] = a[i] + b[j]; ans2[++tot2] = c[i] + d[j]; } } ll ans = 0; // sort(ans1+1,ans1+tot1+1); sort(ans2+1,ans2+tot2+1); for(int i = 1; i<=tot1; i++) { if(binary_search(ans2+1,ans2+tot2+1,-ans1[i])) ans += get(-ans1[i]+1) - get(-ans1[i]); } printf("%lld\n",ans); } return 0 ; }

总结:

  本来想着可能会炸空间,是不是需要离散化一下,但是后来一想发现不可以这样,相反,如果离散化了这题就错了,因为他要计算的是次数,所以不能去重啊!!而且枚举的时候不能直接加binarysearch,而是应该先判断是否存在,如果存在的话加上所有存在的个数,可以按照我这样写,也可以一个lowerbound一个upperbound、。、、、

事实证明  longlong的1e7是开的下的,这次提交一共用了97796kB内存而已、、、。

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