Codeforces 321A

x[i] 表示：从第一步到第i步为止，横坐标的位置；

y[i]表示：从第一步到第i步为止， 纵坐标的位置；

设 字符串S的长度为 len;

则有 k * x[len] + x[i] = a; 而且 k * y[len] + y[i] = b; （这里 下标从1开始），而且要注意的是 k 为非负整数， 还要考虑 x[len] 或 y[len]为0的情况。然后从1到len枚举 i 就可以了。

附上代码：

# by Stomach_ache
def move(s, i):
        if s[i] == 'U':
            return 0, 1;
        elif s[i] == 'D':
            return 0, -1;
        elif s[i] == 'L':
            return -1, 0;
        else:
            return 1, 0;
a, b = map(long, raw_input().split());
s = raw_input();
l = len(s);
x, y = [0]*(l+1), [0]*(l+1)
f = 0;
for i in xrange(1, l+1):
    x[i], y[i] = x[i-1], y[i-1];
    tmpx, tmpy = move(s, i-1)
    x[i] += tmpx;
    y[i] += tmpy;
if a == 0 and b == 0:
    print "Yes";
else:
    # x[l] and y[l] can both be zero, so we can not divide directly
    # k must be a nonegative interger
    for i in xrange(1, l+1):
        if x[l] == 0 and y[l] != 0:
            if a == x[i] and (b - y[i]) % y[l] == 0 and (b - y[i]) / y[l] >= 0:
                f = 1
                break;
        elif y[l] == 0 and x[l] != 0:
            if b == y[i] and (a - x[i]) % x[l] == 0 and (a - x[i]) / x[l] >= 0:
                f = 1
                break;    
        elif x[l] == 0 and y[l] == 0:
            if a == x[i] and b == y[i]:
                f = 1;
                break;
        elif (a - x[i]) % x[l] == 0 and (b - y[i]) % y[l] == 0:
            if (a - x[i]) / x[l] == (b - y[i]) / y[l] and (b - y[i]) / y[l] >= 0:
                f = 1;
                break;

    if f:
        print "Yes";
    else:
        print "No";