文章标题POJ 2785:4 Values whose Sum is 0?(二分)
2016-07-22 08:44
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4 Values whose Sum is 0
DescriptionThe 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).
题意:题目要我们从四个表中各找一个数相加的和为0。
分析:直接暴力破解肯定不行,可先对四个数表分成两组进行枚举相加,然后其中一个枚举,另一个二分,记住,记录重复的情况。
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<math.h> #include<algorithm> using namespace std; int a[4005],b[4005],c[4005],d[4005]; int sum1[4005*4005],sum2[4005*4005]; int bs (int a[],int key,int k){ int lo=0,hi=k-1; int mi,ans=hi+1; while (lo<=hi){ mi=((hi-lo)>>1)+lo; //if (a[mi]==key)return mi; if (a[mi]<key)lo=mi+1; else { hi=mi-1;ans=min(ans,mi); } } if (ans==hi+1) return -1; return ans; } int main () { int n; while (scanf ("%d",&n)!=EOF){ for (int i=0;i<n;i++){ scanf ("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]); } int k1=0; for (int i=0;i<n;i++){ for (int j=0;j<n;j++){ sum1[k1++]=a[i]+b[j]; } } int k2=0; for (int i=0;i<n;i++){ for (int j=0;j<n;j++){ sum2[k2++]=c[i]+d[j]; } } sort (sum1,sum1+k1); int count=0; for (int i=0;i<k2;i++){ count+=upper_bound(sum1,sum1+k1,-sum2[i])-lower_bound(sum1,sum1+k1,-sum2[i]);//点睛之笔 } printf ("%d\n",count); } return 0; }
其中有一个点睛之笔
count+=upper_bound(sum1,sum1+k1,-sum2[i])-lower_bound(sum1,sum1+k1,-sum2[i]);//点睛之笔
这是借鉴别人的成果,通过运用这个,可以减少很多不必要的操作。
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