Infinite String Comparision
链接：https://ac.nowcoder.com/acm/contest/5666/F
来源：牛客网
题目描述
For a string x x x, Bobo defines x ∞ = x x x … x^\infty = x x x \dots x∞=xxx…, which is x x x repeats for infinite times, resulting in a string of infinite length.
Bobo has two strings aa and bb. Find out the result comparing a ∞ a^\infty a∞ and b ∞ b^\infty b∞ in lexicographical order.
You can refer the wiki page for further information of Lexicographical Order.
输入描述:
The input consists of several test cases terminated by end-of-file.
The first line of each test case contains a string a a a, and the second line contains a a a string b b b.
-
1 ≤ ∣ a ∣ , ∣ b ∣ ≤ 1 0 5 1 \leq |a|, |b| \leq 10^5 1≤∣a∣,∣b∣≤105
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a a a, b b b consists of lower case letters.
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The total length of input strings does not exceed 2 × 1 0 6 2 \times 10^6 2×106.
输出描述:
For each test case, print “=” (without quotes) if a ∞ = b ∞ a^\infty = b^\infty a∞=b∞. Otherwise, print “<” if a ∞ < b ∞ a^\infty < b^\infty a∞<b∞, or “>” if a ∞ > b ∞ a^\infty > b^\infty a∞>b∞.
示例1
输入
aa
b
zzz
zz
aba
abaa
输出
<
=
>
设 s 1 = a + b s_1=a+b s1=a+b， s 2 = b + a s_2=b+a s2=b+a，则
s 1 ∞ = a b a b a b … s_1^\infty=ababab\dots s1∞=ababab…
s 2 ∞ = b a b a b a … s_2^\infty=bababa\dots s2∞=bababa…
显然 s 1 ∞ s_1^\infty s1∞和 s 2 ∞ s_2^\infty s2∞都包含子序列 a ∞ a^\infty a∞和 b ∞ b^\infty b∞，且 s 1 s_1 s1， s 2 s_2 s2的字典序大小关系与 s 1 ∞ s_1^\infty s1∞， s 2 ∞ s_2^\infty s2∞的字典序大小关系相同。
令 s 1 ∞ s_1^\infty s1∞， s 2 ∞ s_2^\infty s2∞长度相等，则可以把序列看做是数字，字典序大小看做是数字大小的比较。
比较大小应当是各位的数乘上对应位的权值，不妨设 s 1 ∞ s_1^\infty s1∞中 a ∞ a^\infty a∞对应的权值总和为 s u m 1 sum_1 sum1； b ∞ b^\infty b∞对应的权值总和为 s u m 2 sum_2 sum2，则 s 1 ∞ s_1^\infty s1∞中 a ∞ a^\infty a∞对应的权值总和为 s u m 2 sum_2 sum2； b ∞ b^\infty b∞对应的权值总和为 s u m 1 sum_1 sum1。
s 1 ∞ − s 2 ∞ = ( s u m 1 ⋅ a ∞ + s u m 2 ⋅ b ∞ ) − ( s u m 2 ⋅ a ∞ + s u m 1 ⋅ b ∞ ) = ( s u m 1 − s u m 2 ) ⋅ ( a ∞ − b ∞ ) \begin{aligned} s_1^\infty-s_2^\infty &=(sum_1·a^\infty+sum_2·b^\infty)-(sum_2·a^\infty+sum_1·b^\infty)\\ &=(sum_1-sum_2)·(a^\infty-b^\infty) \end{aligned} s1∞−s2∞=(sum1⋅a∞+sum2⋅b∞)−(sum2⋅a∞+sum1⋅b∞)=(sum1−sum2)⋅(a∞−b∞)
显然 s u m 1 − s u m 2 > 0 sum_1-sum_2>0 sum1−sum2>0
因此 a ∞ a^\infty a∞， b ∞ b^\infty b∞的字典序大小关系与 s 1 ∞ s_1^\infty s1∞， s 2 ∞ s_2^\infty s2∞的字典序大小关系相同。
即 a ∞ a^\infty a∞， b ∞ b^\infty b∞的字典序大小关系与 s 1 s_1 s1， s 2 s_2 s2的字典序大小关系相同。
因此直接比较 s 1 s_1 s1， s 2 s_2 s2的字典序大小关系即可。复杂度为 O ( n ) O(n) O(n)。
#include
using namespace std;
inline int qr() {
int f = 0, fu = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-')fu = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
f = (f << 3) + (f << 1) + c - 48;
c = getchar();
}
return f * fu;
}
string a, b;
string s1, s2;
int main() {
while (cin >> a, cin >> b) {
s1 = a + b, s2 = b + a;
if (s1 > s2)puts(">");
else if (s1 < s2)puts("<");
else puts("=");
}
return 0;
}