[USACO1.3]号码锁 Combination Lock
生活随笔
收集整理的這篇文章主要介紹了
[USACO1.3]号码锁 Combination Lock
小編覺得挺不錯的,現在分享給大家,幫大家做個參考.
https://www.luogu.org/recordnew/show/17460324
題解:
/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp,sum; int psd[2][3]; char str; struct node{}; bool vis[2][3][N]; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);scanf("%d%d%d",&psd[0][0],&psd[0][1],&psd[0][2]);scanf("%d%d%d",&psd[1][0],&psd[1][1],&psd[1][2]);for(int i=0;i<=1;i++){for(int j=0;j<=2;j++){for(int k=psd[i][j]-2;k<=psd[i][j]+2;k++){if(k<1){vis[i][j][k+n]=1;}else if(k>n){vis[i][j][k-n]=1;}else{vis[i][j][k]=1;}}}}for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)for(int k=1;k<=n;k++)if(vis[0][0][i]&&vis[0][1][j]&&vis[0][2][k])ans++;else if(vis[1][0][i]&&vis[1][1][j]&&vis[1][2][k])ans++;cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }?
總結
以上是生活随笔為你收集整理的[USACO1.3]号码锁 Combination Lock的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: [USACO1.3]滑雪课程设计Ski
- 下一篇: [USACO1.4]母亲的牛奶 Moth