uva 1152 4 values whose sum is zero ——yhx
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) 2 AB C D are such that a+b+c+d = 0. In the following, we
assume that all lists have the same size n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below. This line is followed by a blank line, and there is also a
blank line between two consecutive inputs.
The rst line of the input le 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 228) that belong
respectively to A;B;C and D.
Output
For each test case, your program has to write the number quadruplets whose sum is zero.
The outputs of two consecutive cases will be separated by a blank line.
how many quadruplet (a; b; c; d) 2 AB C D are such that a+b+c+d = 0. In the following, we
assume that all lists have the same size n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below. This line is followed by a blank line, and there is also a
blank line between two consecutive inputs.
The rst line of the input le 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 228) that belong
respectively to A;B;C and D.
Output
For each test case, your program has to write the number quadruplets whose sum is zero.
The outputs of two consecutive cases will be separated by a blank line.
1 #include<cstdio> 2 #include<cstring> 3 int abs(int x) 4 { 5 if (x>=0) return x; 6 return -x; 7 } 8 const int m=1098469; 9 int a[4010],b[4010],c[4010],d[4010],first[1100000],next[17000000],num[17000000]; 10 int main() 11 { 12 int i,j,k,n,p,q,x,y,z,t,ans; 13 scanf("%d",&t); 14 while (t--) 15 { 16 memset(a,0,sizeof(a)); 17 memset(b,0,sizeof(b)); 18 memset(c,0,sizeof(c)); 19 memset(d,0,sizeof(d)); 20 memset(first,0,sizeof(first)); 21 memset(next,0,sizeof(next)); 22 memset(num,0,sizeof(num)); 23 ans=0; 24 scanf("%d",&n); 25 for (i=1;i<=n;i++) 26 scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]); 27 for (i=1;i<=n;i++) 28 for (j=1;j<=n;j++) 29 { 30 x=a[i]+b[j]; 31 p=abs(x%m); 32 next[(i-1)*n+j]=first[p]; 33 first[p]=(i-1)*n+j; 34 num[(i-1)*n+j]=x; 35 } 36 for (i=1;i<=n;i++) 37 for (j=1;j<=n;j++) 38 { 39 x=-c[i]-d[j]; 40 p=abs(x%m); 41 for (k=first[p];k;k=next[k]) 42 if (x==num[k]) ans++; 43 } 44 printf("%d\n",ans); 45 if (t) printf("\n"); 46 } 47 }
枚举a+b,把所有值存起来,然后枚举-c-d,在a+b中查找。
具体查找方法是哈希,除k取余法即可。