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  • UVALive 6858 Frame （模拟）

    Frame

    题目链接：

    http://acm.hust.edu.cn/vjudge/contest/130303#problem/D

    Description

    http://7xjob4.com1.z0.glb.clouddn.com/17e6574035df3f9b6d1fc6dfd8b650ac

    Input

    The input file contains several test cases, each of them as described below. The first line contains 2 integers — X and Y (3 ≤ X ≤ 10^6, 3 ≤ Y ≤ 10^6). The second line contains integer N — the number of tile types to be analyzed (1 ≤ N ≤ 1000). Each of following N lines contains one integer, not exceeding 10^6. We designate with AK the integer on the (k + 2)-th line of the input file.

    Output

    For each test case, your goal is to print N lines, where the K-th line should contain the word ‘YES’, if it is possible to tile the frame with size X × Y with tiles AK × 1, and the word ‘NO’ otherwise.

    Sample Input

    ``` 5 6 2 3 4 ```

    Sample Output

    ``` YES NO ```

    Source

    2016-HUST-线下组队赛-4
    ##题意： 在X*Y的矩形的最外圈放若干个 A*1 的小矩形，判断对于给定的A能够恰好完全覆盖.
    ##题解： 考虑最上面一行的情况，第一块小矩形要么从#(1,1)开始，要么从#(1,2)开始(留出第一格给竖的矩形). 确定第一块的位置后，后面的摆放情况是固定的，所以依次判断一下是否合法即可.
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define maxn 101000 #define inf 0x3f3f3f3f #define mod 1000000007 #define mid(a,b) ((a+b)>>1) #define eps 1e-8 #define IN freopen("in.txt","r",stdin); using namespace std;

    int n;

    int x,y;
    bool solve(int a) {
    int last1 = x % a;
    if(last1 > 1) return 0;
    if(last1 == 1) {
    int last2 = y % a;
    if(last2 > 1) return 0;
    if(last2 == 1) return 1;
    if(last2 == 0) return (y-2) % a == 0;
    }
    if(last1 == 0) {
    int last2 = (y-1) % a;
    if(last2 > 1) return 0;
    if(last2 == 1) return 1;
    if(last2 == 0) return (x-2) % a == 0;
    }
    }

    bool solve2(int a) {
    int last1 = (x-1) % a;
    if(last1 > 1) return 0;
    if(last1 == 1) {
    int last2 = y % a;
    if(last2 > 1) return 0;
    if(last2 == 0) return 1;
    if(last2 == 1) return x % a == 0;
    }
    if(last1 == 0) {
    int last2 = (y-1) % a;
    if(last2 > 1) return 0;
    if(last2 == 0) return 1;
    if(last2 == 1) return y % a == 0;
    }
    }

    int main()
    {
    //IN;

    while(scanf("%d %d", &x,&y) != EOF)
{
    swap(x,y);
    int n; scanf("%d", &n);
    while(n--) {
        int a; scanf("%d", &a);
        if(solve(a) || solve2(a)) puts("YES");
        else puts("NO");
    }
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5811188.html
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