1339 - Ancient Cipher

本文介绍了一种基于古罗马加密技术的验证方法,通过对比两个字符串中各字母出现频次来判断是否可通过特定的替换与排列加密方式相互转换。

Ancient Roman Empire had a stronggovernment system with various departments, including a secret servicedepartment. Important documents were sent between provinces and the capital inencrypted form to prevent eavesdropping. The most popular ciphers in thosetimes were so called substitution cipher and permutation cipher.Substitution cipher changes all occurrences of each letter to some otherletter. Substitutes for all letters must be different. For some letterssubstitute letter may coincide with the original letter. For example, applyingsubstitution cipher that changes all letters from `A' to `Y' to the next onesin the alphabet, and changes `Z' to `A', to the message ``VICTORIOUS'' one getsthe message ``WJDUPSJPVT''. Permutation cipher applies some permutation to theletters of the message. For example, applying the permutation2, 1, 5, 4, 3, 7, 6,10, 9, 8 to the message``VICTORIOUS'' one gets the message ``IVOTCIRSUO''. It was quickly noticed thatbeing applied separately, both substitution cipher and permutation cipher wererather weak. But when being combined, they were strong enough for those times.Thus, the most important messages were first encrypted using substitutioncipher, and then the result was encrypted using permutation cipher. Encryptingthe message ``VICTORIOUS'' with the combination of the ciphers described aboveone gets the message ``JWPUDJSTVP''. Archeologists have recently found themessage engraved on a stone plate. At the first glance it seemed completelymeaningless, so it was suggested that the message was encrypted with somesubstitution and permutation ciphers. They have conjectured the possible textof the original message that was encrypted, and now they want to check theirconjecture. They need a computer program to do it, so you have to write one.

Input 

Input file contains several test cases.Each of them consists of two lines. The first line contains the messageengraved on the plate. Before encrypting, all spaces and punctuation marks wereremoved, so the encrypted message contains only capital letters of the Englishalphabet. The second line contains the original message that is conjectured tobe encrypted in the message on the first line. It also contains only capitalletters of the English alphabet. The lengths of both lines of the input fileare equal and do not exceed 100.

Output 

For each test case, print one output line.Output `YES' if the message on the first line of the input file could be theresult of encrypting the message on the second line, or `NO' in the other case.

Sample Input 

JWPUDJSTVP

VICTORIOUS

MAMA

ROME

HAHA

HEHE

AAA

AAA

NEERCISTHEBEST

SECRETMESSAGES

Sample Output 

YES

NO

YES

YES

NO

代码:

#include<iostream>

#include<cstring>

#include<string>

#include<algorithm>

using namespacestd;

string s1,s2;

inttime1[26],time2[26];

int main()

{

    while(getline(cin,s1))

    {

        getline(cin,s2);

        memset(time1,0,sizeof(time1));

        memset(time2,0,sizeof(time2));

        int n=s1.size();

        for(int i=0;i<n;i++)

        {

            time1[s1[i]-'A']++;

            time2[s2[i]-'A']++;

        }

        sort(time1,time1+26);

        sort(time2,time2+26);

        int check=1;

        for(int i=0;i<26;i++)

        {

            if(time1[i]!=time2[i])

            {

               cout<<"NO"<<endl;

                check=0;

                break;

            }

        }

        if(check==1)

        {

           cout<<"YES"<<endl;

        }

    }

}

思路:以两个长度相同的字符串为单位进行比较,用两个int数组分别统计两个字符串中每一种大写字母出现的次数,统计完后,对两个int型数组进行排序,排序结果相同证明可以通过某种映射(千万不要理解错题意,题目中提到的向后映射仅仅是一个例子而已,并非只能采取这种方式的映射,也就是说,映射的方式与规则是未知的,但有一点是确定的:不同字母的映射结果肯定是不同的)将字符串1转换成字符串2,一旦转换成功,再乱排也肯定能满足条件.

static void ath10k_set_key_h_def_keyidx(struct ath10k *ar, struct ath10k_vif *arvif, enum set_key_cmd cmd, struct ieee80211_key_conf *key) { u32 vdev_param = arvif->ar->wmi.vdev_param->def_keyid; int ret; /* 10.1 firmware branch requires default key index to be set to group * key index after installing it. Otherwise FW/HW Txes corrupted * frames with multi-vif APs. This is not required for main firmware * branch (e.g. 636). * * This is also needed for 636 fw for IBSS-RSN to work more reliably. * * FIXME: It remains unknown if this is required for multi-vif STA * interfaces on 10.1. */ if (arvif->vdev_type != WMI_VDEV_TYPE_AP && arvif->vdev_type != WMI_VDEV_TYPE_IBSS) return; if (key->cipher == WLAN_CIPHER_SUITE_WEP40) return; if (key->cipher == WLAN_CIPHER_SUITE_WEP104) return; if (key->flags & IEEE80211_KEY_FLAG_PAIRWISE) return; if (cmd != SET_KEY) return; ret = ath10k_wmi_vdev_set_param(ar, arvif->vdev_id, vdev_param, key->keyidx); if (ret) ath10k_warn(ar, "failed to set vdev %i group key as default key: %d\n", arvif->vdev_id, ret); } static int ath10k_set_key(struct ieee80211_hw *hw, enum set_key_cmd cmd, struct ieee80211_vif *vif, struct ieee80211_sta *sta, struct ieee80211_key_conf *key) { struct ath10k *ar = hw->priv; struct ath10k_vif *arvif = (void *)vif->drv_priv; struct ath10k_sta *arsta; struct ath10k_peer *peer; const u8 *peer_addr; bool is_wep = key->cipher == WLAN_CIPHER_SUITE_WEP40 || key->cipher == WLAN_CIPHER_SUITE_WEP104; int ret = 0; int ret2; u32 flags = 0; u32 flags2; /* this one needs to be done in software */ if (key->cipher == WLAN_CIPHER_SUITE_AES_CMAC || key->cipher == WLAN_CIPHER_SUITE_BIP_GMAC_128 || key->cipher == WLAN_CIPHER_SUITE_BIP_GMAC_256 || key->cipher == WLAN_CIPHER_SUITE_BIP_CMAC_256) return 1; 是这两个函数吗
04-09
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