poj The Clocks 高斯消元

由于数据量不大，所以这题有很多解法。

我用的是高斯消元化为逆矩阵解决的……

代码如下：

#include<stdio.h>

#include<iostream>

using namespace std;

int an[9][10]={

{1,1,0,1,0,0,0,0,0,0},

{1,1,1,0,1,0,0,0,0,0},

{0,1,1,0,0,1,0,0,0,0},

{1,0,0,1,1,0,1,0,0,0},

{1,0,1,0,1,0,1,0,1,0},

{0,0,1,0,1,1,0,0,1,0},

{0,0,0,1,0,0,1,1,0,0},

{0,0,0,0,1,0,1,1,1,0},

{0,0,0,0,0,1,0,1,1,0}

},ans[9];

int main(){

    int n,i,j,k,sum,ma,mb;

    for(i=0;i<9;i++){

        cin>>an[i][9];

        an[i][9]=(4-an[i][9])%4;

    }

    for(i=0;i<9;i++){

        if(an[i][i]==0){

            for(k=i+1;k<9;k++){

                if(an[k][i]){

                    for(j=0;j<=9;j++)

                        swap(an[i][j],an[k][j]);

                    break;

                }

            }

        }

        for(k=i+1;k<9;k++){

            if(an[k][i]){

                ma=an[i][i];

                mb=an[k][i];

                for(j=i;j<=9;j++){

                    an[k][j]=ma*an[k][j]-mb*an[i][j];

                    an[k][j]=(an[k][j]%4+4)%4;

                }

            }

        }

    }

    sum=0;

    for(i=8;i>=0;i--){

        for(j=i+1;j<9;j++){

            an[i][9]-=ans[j]*an[i][j];

            an[i][9]=(an[i][9]%4+4)%4;

        }

        for(ans[i]=0;ans[i]<=3;ans[i]++)

            if((ans[i]*an[i][i]%4+4)%4==an[i][9])

                break;

        ans[i]%=4;

        sum+=ans[i];

    }

    j=0;k=0;

    while(j<9){

        if(ans[j]){

            cout<<j+1;

            k++;

            if(k>=sum) cout<<endl;

            else cout<<' ';

            ans[j]--;

        }

        else j++;

    }

    return 0;

}

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