Eqs
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 15010 | Accepted: 7366 |
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.
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
按要求模拟即可
hash是个神奇的东西
下方代码注释部分是先三层循环后二层，正文部分是先二层循环后三层。两者都是正解，但是由于list插入比读取慢,先二层更快
1 /* 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<algorithm> 6 #include<list> 7 using namespace std; 8 const int mxn=14997; 9 list<int>ha[mxn*2]; 10 list<int>::iterator it; 11 int x,x1,x2,x3,x4,x5; 12 int ans=0; 13 int main(){ 14 scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5); 15 int i,j,k; 16 for(i=-50;i<=50;i++) 17 for(j=-50;j<=50;j++) 18 for(k=-50;k<=50;k++){ 19 if(i==0||j==0||k==0)continue; 20 x=i*i*i*x1+j*j*j*x2+k*k*k*x3; 21 ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数 22 } 23 for(i=-50;i<=50;i++) 24 for(j=-50;j<=50;j++){ 25 if(i==0|j==0)continue; 26 x=-(i*i*i*x4+j*j*j*x5); 27 //检查hash 28 for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){ 29 if(*it==x)ans++; 30 } 31 } 32 printf("%d",ans); 33 return 0; 34 35 } 36 */ 37 #include<iostream> 38 #include<cstdio> 39 #include<cstring> 40 #include<algorithm> 41 #include<list> 42 using namespace std; 43 const int mxn=14997; 44 list<int>ha[mxn*2]; 45 list<int>::iterator it; 46 int x,x1,x2,x3,x4,x5; 47 int ans=0; 48 int main(){ 49 scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5); 50 int i,j,k; 51 for(i=-50;i<=50;i++) 52 for(j=-50;j<=50;j++) 53 { 54 if(i==0||j==0)continue; 55 x=i*i*i*x1+j*j*j*x2; 56 ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数 57 } 58 for(i=-50;i<=50;i++) 59 for(j=-50;j<=50;j++) 60 for(k=-50;k<=50;k++){ 61 if(i==0|j==0||k==0)continue; 62 x=-(i*i*i*x3+j*j*j*x4+k*k*k*x5); 63 //检查hash 64 for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){ 65 if(*it==x)ans++; 66 } 67 } 68 printf("%d",ans); 69 return 0; 70 71 }
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