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  • POJ 1840 Eqs

    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. 

    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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  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5550601.html
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