Problem E: A multiplication game
时间： 1ms 内存：128M
描述：
Problem E: A multiplication game
Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n. Each line of input contains one integer number n. For each line of input output one line either
Stan wins.or
Ollie wins.assuming that both of them play perfectly.
输入：
输出：
示例输入：
162
17
34012226
示例输出：
Stan wins.
Ollie wins.
Stan wins.
提示：
参考答案（内存最优[748]）：
#include <stdio.h>
#include <assert.h>
int main () {
unsigned int W, l ,h, n;
while (scanf("%u",&n)==1) {
assert(1 < n && n < 4294967295U);
W = 1;
l = n/9; if (l*9<n) l++; h = n-1;
while (1) {
if (l <= 1 && 1 <= h) {
printf(W?"Stan wins.\n":"Ollie wins.\n");
break;
}
if (W) {
if (l/2*2 < l) l = l/2+1; else l/=2;
h/=9;
W=0;
} else {
if (l/9*9 < l) l = l/9+1; else l/=9;
h/=2;
W=1;
}
}
}
return 0;
}参考答案（时间最优[1]）：
#include <stdio.h>
#include <assert.h>
int main () {
unsigned int W, l ,h, n;
while (scanf("%u",&n)==1) {
assert(1 < n && n < 4294967295U);
W = 1;
l = n/9; if (l*9<n) l++; h = n-1;
while (1) {
if (l <= 1 && 1 <= h) {
printf(W?"Stan wins.\n":"Ollie wins.\n");
break;
}
if (W) {
if (l/2*2 < l) l = l/2+1; else l/=2;
h/=9;
W=0;
} else {
if (l/9*9 < l) l = l/9+1; else l/=9;
h/=2;
W=1;
}
}
}
return 0;
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