2017-6-8

題目描述

給你n盞開著的燈，開關(guān)的次數(shù)，以及部分燈最后的狀態(tài)，求出所有可能的情況

解答

史上最長(zhǎng)的代碼 同一個(gè)按鈕按奇數(shù)次都是一樣的，按偶數(shù)次也都是一樣的

代碼

#include<iostream> #include<cstring> #include<cstdio> using namespace std;const int N = 100; char x[N+1]; int f[N+1]; int a[4]; int n,c; char y[N+1][N+1]; int k=0;void init(){int m;cin>>m;while (m!=-1){f[m]=1;cin>>m;}cin>>m;while (m!=-1){f[m]=-1;cin>>m;} }int cal(){int sum=0;for (int i=0;i<4;i++){if (a[i]==1) sum++;}return sum; }void re(){if (a[0]==1){for (int i=1;i<=n;i++){x[i]=(x[i]-'0'+1)%2+'0';}}if (a[1]==1){for (int i=1;i<=n;i++){if (i%2==1) x[i]=(x[i]-'0'+1)%2+'0';}}if (a[2]==1){for (int i=1;i<=n;i++){if (i%2==0) x[i]=(x[i]-'0'+1)%2+'0';}}if (a[3]==1){for (int i=1;i<=n;i++){if (i%3==1) x[i]=(x[i]-'0'+1)%2+'0';}} }bool tru(){for (int i=1;i<=n;i++){if (f[i]==1){if (x[i]!='1') return false;}else if (f[i]==-1){if (x[i]!='0') return false;}}return true; }void set(){for(int i=1;i<=n;i++) x[i]='1'; }void res(){for (a[0]=0;a[0]<=1;a[0]++){for (a[1]=0;a[1]<=1;a[1]++){for (a[2]=0;a[2]<=1;a[2]++){for (a[3]=0;a[3]<=1;a[3]++){set();if (c==0){if (cal()==0){if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}}else if (c==1){if (cal()==1){re();if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}}else if (c==2){if (cal()==0){if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}else if (cal()==2){re();if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}}else{if (c%2==0){if (cal()==0){if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}else if (cal()==2){re();if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}else if (cal()==4){re();if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}}else{if (cal()==1){re();if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}else if (cal()==3){re();if (tru()){for (int i=0;i<n;i++) y[k][i]=x[i+1];k++;}}}}}}}} } void out(){if (k==0) cout<<"IMPOSSIBLE"<<endl;int i,j;for (i=0;i<k;i++) y[i][n]='\0';for (i=0;i<k;i++){}char t[N+1];for (i=0;i<k-1;i++){for (j=0;j<k-i-1;j++){if (strcmp(y[j],y[j+1])>0){strcpy(t,y[j]);strcpy(y[j],y[j+1]);strcpy(y[j+1],t);}}}for (i=0;i<k;i++){cout<<y[i]<<endl;} }int main(){cin>>n>>c;init();res();out(); }

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