UVA - 1435 Business Cards （數論）

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Description

Running a paper shop is not an easy job, especially with harsh customers. Today they brought their own rectangular sheets of paper, asking you to cut it into rectangular business cards of specific size. Moreover, they require that all the paper (which may not be cheap, but is definitely not that expensive!) has to be used, i.e. no tiny bit may be left over. Moreover, the brilliant idea of cutting the sheet into very small pieces, and then gluing them together in desired sheets was laughed at.

An example of a 9×6 paper sheet divided into 2×3 cards is given below.

Input

The input contains several test cases. The first line contains the number of test casest (t105). Thent test cases follow. Each of them consists of one line containing four integersa, b, c, d (1a,b, c, d109). Numbersa and b are dimensions of each business card;c and d are dimensions of the paper sheet.

Output

For each test case output one line containing word `YES‘ if it is possible to divide the whole sheet into business cards, and `NO‘ otherwise.

Sample Input

4 2 3 9 62 3 8 62 3 6 82 3 5 7

Sample Output

YES YES YES NO題意：給你a,b,c,d四個數，問你能把c*d的矩陣分成若干個a*b的矩陣思路：分情況討論：1全都豎著放；2全都橫著放，3交錯放，注意第3種情況只能是長或者是寬是組合的，等於說就是求解ax+by=c|d的這兩種情況，這個可以試著畫一下。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>typedef long long ll;using namespace std;int find(ll a, ll b, ll c) {if (a < b)swap(a, b);for (ll i = a; i < c; i += a)if ((c - i) % b == 0)return 1;return 0;}int main() {ll a, b, c, d;int t;scanf("%d", &t);while (t--) {scanf("%lld%lld%lld%lld", &a, &b, &c, &d);if ((c % a == 0 && d % b == 0) || (d % a == 0 && c % b == 0))printf("YES\n");else if (c % a == 0 && c % b == 0 && find(a, b, d))printf("YES\n");else if (d % a == 0 && d % b == 0 && find(a, b, c))printf("YES\n");else printf("NO\n");}return 0;}

UVA - 1435 Business Cards （數論）