ZOJ-The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple-A~B

ACM模版

心血来潮，参加了这么啥比赛，名字好长啊，只是参加了网赛，感觉还行，就是看不懂题~~~前两题过于简单，所以写在一起吧！

A-Cooking Competition

描述

题解

水题。

代码

#include <iostream>using namespace std;const int K[] = {0, 1, 0, 1, -1};const int D[] = {0, 0, 1, 1, -1};int main(int argc, const char * argv[]){    int T;    cin >> T;    int n, x, resK, resD;    while (T--)    {        cin >> n;        resK = resD = 0;        for (int i = 0; i < n; i++)        {            cin >> x;            resK += K[x];            resD += D[x];        }        if (resK > resD)        {            cout << "Kobayashi\n";        }        else if (resK < resD)        {            cout << "Tohru\n";        }        else        {            cout << "Draw\n";        }    }    return 0;}

B-Problem Preparation

描述

题解

这个题有一个暗坑，明明说难度最低为1，数据范围还 −1000≤si≤1000，所以这里需要注意，如果出现负数，直接 NO。我就是这里没有翻译对，导致 WA 了好几遍。

代码

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 110;int s[MAXN];int main(int argc, const char * argv[]){//    freopen("/Users/zyj/Desktop/input.txt", "r", stdin);    int T;    scanf("%d", &T);    int n;    while (T--)    {        scanf("%d", &n);//        int cnt = 0;        for (int i = 0; i < n; i++)        {            scanf("%d", s + i);//            if (s[i] == 1)//            {//                cnt++;//            }        }        if (n < 10 || n > 13/* || cnt < 2*/)        {            printf("No\n");            continue;        }        sort(s, s + n);        //  ???        if (s[0] != 1 || s[0] != 1)        {            printf("No\n");            continue;        }        int flag = 1;        for (int i = 1; i < n; i++)        {            if (s[i] - s[i - 1] > 2 && s[i] != s[n - 1])            {                flag = 0;                break;            }        }        if (flag)        {            printf("Yes\n");        }        else        {            printf("No\n");        }    }    return 0;}