XidianOJ 1035 数独 && 1053 正数负数 && 1042 另一个简单的游戏
Posted TOTOTOTOTZZZZZ
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了XidianOJ 1035 数独 && 1053 正数负数 && 1042 另一个简单的游戏相关的知识,希望对你有一定的参考价值。
三道水题。。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n; int main(){ while (scanf("%d",&n) != EOF){ if (n > 0){ printf("yes\n"); } else if (n < 0){ printf("no\n"); } else printf("light\n"); } return 0; }
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define SIZE 9 int table[10][10]; bool JudgeCube(int x,int y){ int incx[10] = {0,0,0,0,1,1,1,2,2,2}; int incy[10] = {0,0,1,2,0,1,2,0,1,2}; int i,j; for (i=1;i<=9;i++){ int appear[10] = {0}; int nowx = x + incx[i]; int nowy = y + incy[i]; if (appear[table[nowx][nowy]]) { return true;; } else appear[table[nowx][nowy]]; } return false; } int main(){ int time,T; scanf("%d",&T); for (time=1;time<=T;time++){ int i,j; int ok = 1; for (i=1;i<=SIZE;i++){ for (j=1;j<=SIZE;j++){ scanf("%d",&table[i][j]); } } for (i=1;i<=SIZE;i++){ int appear[10] = {0}; for (j=1;j<=SIZE;j++){ if (table[i][j] > 9 || table[i][j] < 1) { ok = 0; goto here; } if (appear[table[i][j]]) { ok = 0; goto here; } else appear[table[i][j]] = 1; } } for (i=1;i<=SIZE;i++){ int appear[10] = {0}; for (j=1;j<=SIZE;j++){ if (appear[table[j][i]]) { ok = 0; goto here; } else appear[table[j][i]] = 1; } } if (JudgeCube(1,1) || JudgeCube(1,4) || JudgeCube(1,7) || JudgeCube(4,1) || JudgeCube(4,4) || JudgeCube(4,7) || JudgeCube(7,1) || JudgeCube(7,4) || JudgeCube(7,7)) ok = 0; here: if (ok) { printf("yes\n"); } else printf("no\n"); } return 0; }
#include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <vector> using namespace std; priority_queue<double,vector<double>,greater<double> > q; int main(){ int time,T; scanf("%d",&T); for (time=1;time<=T;time++){ int n; scanf("%d",&n); int i,j; while (!q.empty()) { q.pop(); } for (i=0;i<n;i++){ double element; scanf("%lf",&element); q.push(element); } while (!q.empty()) { double min1 = q.top(); q.pop(); double min2 = q.top(); q.pop(); double res = (min1+min2) / 2; if (q.empty()) { res *= 100; res += 0.5; res = floor(res); res /= 100; printf("%.2lf\n",res); break; } q.push(res); } } return 0; }
以上是关于XidianOJ 1035 数独 && 1053 正数负数 && 1042 另一个简单的游戏的主要内容,如果未能解决你的问题,请参考以下文章