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

hdu1035

POJ1035&&POJ3080&&POJ1936

一波水题 MZOJ 1035: 贝克汉姆

poj 1035(水题,注意增删时的操作)

PAT 甲级 A1035 (2019/02/10)

luogu 1035