Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents were sent between provinces and the capital in encrypted
form to prevent eavesdropping. The most popular ciphers in those times were so called substitution cipher and permutation
cipher. Substitution cipher changes all occurrences of each letter to some other letter. Substitutes for all letters must be different. For some letters substitute letter may coincide with
the original letter. For example, applying substitution cipher that changes all letters from `A' to `Y'
to the next ones in the alphabet, and changes `Z' to `A', to the message ``VICTORIOUS''
one gets the message ``WJDUPSJPVT''. Permutation cipher applies some permutation to the letters of the message. For example, applying the permutation 2,
1, 5, 4, 3, 7, 6, 10, 9, 8
to the message ``VICTORIOUS''
one gets the message ``IVOTCIRSUO''. It was quickly noticed that being applied separately, both substitution cipher and permutation cipher were rather weak. But when being combined,
they were strong enough for those times. Thus, the most important messages were first encrypted using substitution cipher, and then the result was encrypted using permutation cipher. Encrypting the message ``VICTORIOUS''
with the combination of the ciphers described above one gets the message ``JWPUDJSTVP''. Archeologists have recently found the message engraved on a stone plate. At the first glance
it seemed completely meaningless, so it was suggested that the message was encrypted with some substitution and permutation ciphers. They have conjectured the possible text of the original message that was encrypted, and now they want to check their conjecture.
They need a computer program to do it, so you have to write one.
题意:给定2个字符串,判断做一个映射后能否一一对应。
思路:把每个字母个数保存下来,排序,看能否一一对应即可
代码:
#include#include #include using namespace std; const int N = 105; char str[N]; int vis1[26], vis2[26]; bool judge() { for (int i = 0; i < 26; i ++) if (vis1[i] != vis2[i]) return false; return true; } int main() { while (~scanf("%s", str)) { memset(vis1, 0, sizeof(vis1)); memset(vis2, 0, sizeof(vis2)); for (int i = 0; i < strlen(str); i ++) vis1[str[i] - 'A'] ++; sort(vis1, vis1 + 26); scanf("%s", str); for (int i = 0; i < strlen(str); i ++) vis2[str[i] - 'A'] ++; sort(vis2, vis2 + 26); if (judge()) printf("YES\n"); else printf("NO\n"); } return 0; }