Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?
InputFirst line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.OutputFor each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".Sample Input
2 2 1 -1 1 -1 1 -1 1 -1 1 -1 3 1 2 3 -1 -2 -3 4 5 6 -1 3 2 -4 -10 -1
Sample Output
No Yes
给你5个数组每个都有N个元素，然后在每一个数组里面取一个数看能否为0
这题5个for不用想肯定TEL
分治的思想 将第一个数组和第二个数组合并 第三个和第四个合并
这样就只有3个数组了。
下面的上代码 ，细节在代码里面体现
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 long long cnt[6][205]; 6 long long a[40005],b[40005],c[40005]; 7 8 int main() { 9 int t; 10 scanf("%d",&t); 11 while(t--){ 12 int n; 13 scanf("%d",&n); 14 for (int i=0 ;i<5 ;i++ ){ 15 for (int j=0 ;j<n ;j++){ 16 scanf("%lld",&cnt[i][j]); 17 } 18 } 19 int k=0; 20 for (int i=0 ;i<n ;i++ ){ 21 for (int j=0 ;j<n ;j++ ){ 22 a[k]=cnt[1][i]+cnt[2][j]; 23 k++; 24 } 25 } 26 k=0; 27 for (int i=0 ;i<n ;i++ ){ 28 for (int j=0 ;j<n ;j++ ){ 29 b[k]=cnt[3][i]+cnt[4][j]; 30 k++; 31 } 32 } 33 int len=0; 34 for (int i=0 ;i<n ;i++){ 35 c[len++]=cnt[0][i]; 36 } 37 sort(a,a+k); 38 sort(b,b+k); 39 sort(c,c+len); 40 int flag=1,temp1,temp2; 41 for (int i=0 ;i<n ;i++){ 42 temp1=k-1; 43 temp2=0; 44 while(temp1>=0 &&temp2<k) { 45 if (a[temp1]+b[temp2]+c[i]==0) { 46 flag=0; 47 break; 48 }else if (a[temp1]+b[temp2]+c[i]>0) { 49 temp1--; 50 }else temp2++; 51 } 52 if (flag==0) break; 53 } 54 if (flag==0) printf("Yes\n"); 55 else printf("No\n"); 56 } 57 return 0; 58 }
