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hdu 1469 Equations
Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5226 Accepted Submission(s): 2079
Problem Description
Consider equations having the following form:
a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.
Determine how many solutions satisfy the given equation.
a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.
Determine how many solutions satisfy the given equation.
Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
End of file.
End of file.
Output
For each test case, output a single line containing the number of the solutions.
Sample Input
1 2 3 -4 1 1 1 1
Sample Output
39088 0
哈希表基础题:
#include<stdio.h> #include<algorithm> #define M 2000004 int hash[M]; int main () { int a,b,c,d; int i,j; int sum,s[101]; for(i=1;i<=100;i++) s[i]=i*i; while(~scanf("%d%d%d%d",&a,&b,&c,&d)) { if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0)) { printf("0\n"); continue; } sum=0; memset(hash,0,sizeof(hash)); for(i=1;i<=100;i++) for(j=1;j<=100;j++) hash[a*s[i]+b*s[j]+M/2]++; // M/2 是防止下标为负数。 for(i=1;i<=100;i++) for(j=1;j<=100;j++) sum+=hash[-(c*s[i]+d*s[j])+M/2]; // 亏大神们想得出啊! printf("%d\n",sum*16); } return 0; }
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