POJ 1840 Eqs
2012-12-12 23:10
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Eqs
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 9567 | Accepted: 4710 |
Consider equations having the following form:
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0
The coefficients are given integers from the interval [-50,50].
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.
Determine how many solutions satisfy the given equation.
Input
The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.
Output
The output will contain on the first line the number of the solutions for the given equation.
Sample Input
37 29 41 43 47
Sample Output
654
题解:简单哈希。数组定义成char型。
#include<iostream> #include<cstring> #include<cstdio> using namespace std; const int MOD=19999999; int ans[111]; char hashtb[MOD]; int main() { int n,cnt=0; int a1,a2,a3,a4,a5; for(int i=-50;i<=50;i++){ if(i==0)continue; ans[cnt++]=i*i*i; } while(scanf("%d %d %d %d %d",&a1,&a2,&a3,&a4,&a5)!=EOF){ memset(hashtb,0,sizeof(hashtb)); for(int i=0;i<100;i++){ for(int j=0;j<100;j++) for(int k=0;k<100;k++){ int temp=a1*ans[i]+a2*ans[j]+a3*ans[k]; if(temp<0) temp+=MOD; temp=temp%MOD; hashtb[temp]++; } } cnt=0; for(int i=0;i<100;i++){ for(int j=0;j<100;j++){ int temp=-(a4*ans[i]+a5*ans[j]); if(temp<0) temp+=MOD; temp=temp%MOD; cnt+=hashtb[temp]; } } printf("%d\n",cnt); } return 0; }
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