POJ题目1840 Eqs(hash)
2015-02-26 22:57
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Eqs
Description
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
Sample Output
Source
Romania OI 2002
用int超内存
ac代码
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 12949 | Accepted: 6349 |
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
Source
Romania OI 2002
用int超内存
ac代码
#include<stdio.h> #include<string.h> short hash[25000020]; int sum; int a,b,c,d,e; void sol() { int i,j,k; memset(hash,0,sizeof(hash)); for(i=-50;i<=50;i++) for(j=-50;j<=50;j++) for(k=-50;k<=50;k++) { if(i==0||j==0||k==0) continue; int res=a*i*i*i+b*j*j*j+c*k*k*k; if(res<0) res+=25000000; hash[res]++; } } void fun() { int i,j; for(i=-50;i<=50;i++) for(j=-50;j<=50;j++) { if(i==0||j==0) continue; int res=d*i*i*i+e*j*j*j*-1; if(res<0) res=25000000+res; sum+=hash[res]; } } int main() { // int a,b,c,d,e; while(scanf("%d%d%d%d%d",&a,&b,&c,&d,&e)!=EOF) { int i,j,k; sum=0; sol(); fun(); printf("%d\n",sum); } }
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