CodeForces 321A Ciel and Robot（数学模拟）

题目链接：http://codeforces.com/problemset/problem/321/A

题意：在一个二维平面中，开始时在（0,0）点，目标点是（a，b），问能不能通过重复操作题目中的指令，从原点移动到目标点。

分析：假设一次完成所有的命令后，移动到了（xx，yy），并且从（Xi，Yi）重复操作k次指令到达目标点，则可以列出方程 Xi + k * xx = a && Yi + k * yy = b，然后解出k，判断k是否大于等于0即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef __int64 LL;
LL a, b, xx, yy;
LL X[150], Y[150];
int len;
bool check(LL x, LL y) {LL tmp_x = a - x, tmp_y = b - y;if(xx == 0) {if(yy == 0) {if(tmp_x == 0 && tmp_y == 0) return true;else return false;}else {if(tmp_y % yy == 0) {if(tmp_y / yy >= 0 && tmp_x == 0) return true;else return false;}else return false;}}else {if(yy == 0) {if(tmp_x % xx == 0) {if(tmp_x / xx >= 0 && tmp_y == 0) return true;else return false;}else return false;}else {if(tmp_x % xx == 0 && tmp_y % yy == 0) {if(tmp_x / xx >= 0 && tmp_y / yy >= 0 && tmp_x / xx == tmp_y / yy) return true;else return false;}else return false;}}
}int main() {char op[150];while(~scanf("%I64d%I64d", &a, &b)) {scanf("%s", op);if(a == 0 && b == 0) {printf("Yes\n");continue;}len = strlen(op);LL x = 0, y = 0;int FLAG = 0;for(int i = 0; i < len; i++) {if(op[i] == 'U') y++;else if(op[i] == 'D') y--;else if(op[i] == 'L') x--;else if(op[i] == 'R') x++;X[i] = x, Y[i] = y;}xx = X[len-1], yy = Y[len-1];for(int i = 0; i < len; i++) {if(check(X[i], Y[i])) {FLAG = 1;printf("Yes\n");break;}}if(!FLAG)  printf("No\n");}return 0;
}

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