給@蕾米莉亞_斯卡雷特的代碼演示

對(duì)于av號(hào)轉(zhuǎn)bv號(hào)/bv號(hào)轉(zhuǎn)av號(hào) c語(yǔ)言一文中的代碼

如何用30行代碼實(shí)現(xiàn)幾乎相同的功能

#include<stdio.h>

#include<stdlib.h>

int to_a[256],to_b[256],check_res,order[]= {9,8,1,6,2,4,0,7,3,5},data[]={13,12,46,31,43,18,40,28,5,54,20,15,8,39,57,45,36,38,51,42,49,52,53,7,4,9,50,10,44,34,6,25,1,26,29,56,3,24,0,47,27,22,41,16,11,37,2,35,21,17,33,30,48,23,55,32,14,19};

char input[13],cset[] ={"123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"};

int is_digit(char *s) {

for(char* ss=s; *ss; ss++)if('0'>*ss||*ss>'9')return 0;

return 1;

}//判斷是不是純數(shù)字(av號(hào))?

unsigned long long B_to_A(char* s) {

unsigned long long sum = 0;

for(int i=9; i>=0; i--)sum*=58,sum+=to_a[s[order[i]]];//order控制順序?

return (sum-100618342136696320)^0xa93b324;//這個(gè)異或等價(jià)于那十幾行零一串的代碼?

}//轉(zhuǎn)化為av號(hào)的等價(jià)代碼

char* A_to_B(char* s) {

unsigned long long? av = (strtoull(s,NULL,10)^0xa93b324)+100618342136696320;

for(int i=0; i<10; i++)s[order[i]]=to_b[av%58],av/=58;//直接和上面反過(guò)來(lái)?

return s;

}//轉(zhuǎn)化為Bv號(hào)的等價(jià)代碼?

int main() {

for(int i=0; i<58; i++) {

to_a[cset[i]] = data[i];

to_b[data[i]] = cset[i];//直接用數(shù)組索引建立雙射?

}

printf("請(qǐng)輸入av/bv號(hào)：");

gets(input);//這里有一點(diǎn)不同的是,要求格式必須正確,然后去掉前兩個(gè)字母?

if(!is_digit(input+2))printf("您輸入的bv號(hào)所對(duì)應(yīng)的av號(hào)是：av%llu",B_to_A(input+2));

else printf("您輸入的av號(hào)所對(duì)應(yīng)的bv號(hào)是：BV%s",A_to_B(input+2));

}

????????這里提出幾個(gè)思想，一是用數(shù)組保存加密的結(jié)果，建立映射，用索引直接訪問(wèn)，避免使用函數(shù)遍歷（其實(shí)這樣也更快了），然后個(gè)數(shù)位轉(zhuǎn)換和順序可以通過(guò)簡(jiǎn)單邏輯和循環(huán)實(shí)現(xiàn)。對(duì)于無(wú)符號(hào)長(zhǎng)整形可以直接使用位運(yùn)算（異或運(yùn)算）以避免轉(zhuǎn)化成二進(jìn)制。

????????如果加入安全輸入，即判斷輸入的格式是否合法，代碼會(huì)變成60行（其實(shí)基本功能也就十多行，大部分還是輸入和判斷）。

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<ctype.h>

const int len = 58;

int to_a[256],to_b[256],check_res,order[]= {9,8,1,6,2,4,0,7,3,5},

? ? ? ? data[]= {13,12,46,31,43,18,40,28,5,54,20,15,8,39,57,45,36,38,51,42,49,52,53,7,4,9,50,10,44,34,6,25,1,26,29,56,3,24,0,47,27,22,41,16,11,37,2,35,21,17,33,30,48,23,55,32,14,19};

char input[13],cset[] = {"123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"};

int is_digit(char *s) {

????for(char* ss=s; *ss; ss++)if(!isdigit(*ss))return 0;

return 1;

}

int is_Bv(char *s) {

for(char* ss=s; *ss; ss++)if(!to_a[*ss])return 0;return 1;

}

int checkavOrBv(char* s) {

if(!*s)return -1;

char hat[13];

strncpy(hat,s,2);

strupr(hat);

if(!strcmp(hat,"av")) {

strcpy(s,s+2);

return is_digit(s)?1:-1;

} else if(!strcmp(hat,"BV")) {

strcpy(s,s+2);

return is_Bv(s)?0:-1;

} else return is_digit(s)?1:is_Bv(s)?0:-1;

}

unsigned long long B_to_A(char* s) {

unsigned long long sum = 0;

for(int i=9; i>=0; i--)sum*=len,sum+=to_a[s[order[i]]];

return (sum-100618342136696320)^0xa93b324;

}

char* A_to_B(char* s) {

unsigned long long? av = (strtoull(s,NULL,10)^0xa93b324)+100618342136696320;

printf("%llu",strtoull(s,NULL,10));

for(int i=0; i<10; i++)s[order[i]]=to_b[av%len],av/=len;s[10]=0;

return s;

}

int main() {

for(int i=0; i<len; i++) {

to_a[cset[i]] = data[i];

to_b[data[i]] = cset[i];

}

printf("請(qǐng)輸入av/bv號(hào)：");

while(true) {

gets(input);

if((check_res=checkavOrBv(input))==-1) {

printf("你輸入的格式不正確,請(qǐng)重新輸入：");

continue;

} else if(check_res==0)printf("您輸入的bv號(hào)所對(duì)應(yīng)的av號(hào)是：av%llu",B_to_A(input));

else printf("您輸入的av號(hào)所對(duì)應(yīng)的bv號(hào)是：BV%s",A_to_B(input));

break;

}

}

在av號(hào)超過(guò)0xFFFFFFF(十六進(jìn)制)后兩者結(jié)果會(huì)不一樣，是正?，F(xiàn)象，現(xiàn)在的av號(hào)也不會(huì)超過(guò)這個(gè)值。至于為什么不一樣，可以自己思考一下。