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将M进制的数X转换为N进制的数输出。
输入的第一行包括两个整数：M和N(2<=M,N<=36)。
下面的一行输入一个数X，X是M进制的数，现在要求你将M进制的数X转换成N进制的数输出。
输出X的N进制表示的数。
16 10 F
15
输入时字母部分为大写，输出时为小写，并且有大数据。
这题初看起来另我头疼，考虑不难但是很麻烦
一开始想的比较绕，准备先把任意进制转换成10进制，再转化成n进制，于是写出了下面的代码
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <iostream> 5 6 using namespace std; 7 8 char mnum[1002]; 9 int nnum[1002]; 10 int tnCnt; 11 12 int tnum[1002]; 13 int tCnt; 14 15 int tm[1002]; 16 int tmCnt; 17 18 int tSum[1002]; 19 int tsCnt; 20 21 int tDiv[1002]; 22 int tdCnt; 23 24 int m, n; 25 26 int getV(char a) { 27 if(a >= ‘0‘ && a <= ‘9‘) { 28 return a - ‘0‘; 29 } 30 else if(a >= ‘a‘ && a <= ‘z‘) { 31 return a - ‘a‘ + 10; 32 } 33 else if(a >= ‘A‘ && a <= ‘Z‘) { 34 return a - ‘A‘ + 10; 35 } 36 } 37 38 char putV(int a) { 39 if(a >= 0 && a <= 9) { 40 return a + ‘0‘; 41 } 42 else { 43 return a - 10 + ‘a‘; 44 } 45 } 46 47 void multiTm() { 48 int ci = 0; 49 int j = 0; 50 for(int i = 0; i < tmCnt; i++) { 51 int ben = tm[i] * m + ci; 52 tm[j++] = ben % 10; 53 ci = ben/10; 54 } 55 if(ci != 0) { 56 tm[j++] = ci; 57 } 58 tmCnt = j; 59 } 60 61 int multi(int a[], int acnt, int b[], int t) { 62 int ci = 0; 63 int j = 0; 64 for(int i = 0; i < acnt; i++) { 65 int ben = a[i] * t + ci; 66 b[j++] = ben % 10; 67 ci = ben/10; 68 } 69 if(ci != 0) { 70 b[j++] = ci; 71 } 72 return j; 73 } 74 75 int add(int a[], int acnt, int b[], int bcnt) { 76 int p = max(acnt, bcnt); 77 int j = 0; 78 int ci = 0; 79 for(int i = 0; i < p; i++) { 80 int ben = a[i] + b[i] + ci; 81 a[j++] = ben % 10; 82 ci = ben/10; 83 } 84 if(ci != 0) { 85 a[j++] = ci; 86 } 87 return j; 88 } 89 90 int div(int t) { 91 int j = 0; 92 int ci = 0; 93 for(int i = 0; i < tdCnt; i++) { 94 int ben = ci*10 + tDiv[i]; 95 tDiv[j++] = ben/t; 96 ci = ben%t; 97 //printf("%d\n",ci); 98 } 99 int state = 0; 100 int p = 0; 101 for(int i = 0; i < j; i++) { 102 if(state == 0 && tDiv[i] != 0) { 103 state = 1; 104 tDiv[p++] = tDiv[i]; 105 } 106 else if(state == 1) { 107 tDiv[p++] = tDiv[i]; 108 } 109 } 110 tdCnt = p; 111 112 /*for(int i = 0; i <= tdCnt; i++) { 113 printf("%d",tDiv[i]); 114 } 115 puts("");*/ 116 117 return ci; 118 } 119 120 int main(int argc, char const *argv[]) 121 { 122 while(scanf("%d %d",&m,&n) != EOF) { 123 scanf("%s",mnum); 124 //to-ten 125 int lenm = strlen(mnum); 126 127 memset(tm, 0, sizeof(tm)); 128 tm[0] = 1; 129 tmCnt = 1; 130 131 memset(tnum, 0, sizeof(tnum)); 132 tCnt = 0; 133 134 for(int i = lenm -1; i >= 0; i--) { 135 int tmp = getV(mnum[i]); 136 memset(tSum, 0, sizeof(tSum)); 137 tsCnt = 0; 138 tsCnt = multi(tm, tmCnt, tSum, tmp); 139 tCnt = add(tnum,tCnt,tSum,tsCnt); 140 multiTm(); 141 } 142 143 /*for(int i = tCnt-1; i >= 0; i--) { 144 printf("%d",tnum[i]); 145 } 146 puts("");*/ 147 for(int i = tCnt-1; i >= 0; i--) { 148 tDiv[tCnt-i-1] = tnum[i]; 149 } 150 tdCnt = tCnt; 151 int j = 0; 152 memset(nnum, 0, sizeof(nnum)); 153 while(!(tdCnt == 0 && tDiv[0] == 0)) { 154 nnum[j++] = div(n); 155 } 156 for(int i = j-1; i >= 0; i--) { 157 printf("%c",putV(nnum[i])); 158 } 159 puts(""); 160 } 161 return 0; 162 }
虽然例子能跑出来，但提交答案错误
后来一拍脑门，进制直接除n取余就好了，转化成10进制简直多此一举，于是写出下面代码
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <iostream> 5 6 using namespace std; 7 8 char mnum[1002]; 9 10 11 int tDiv[1002]; 12 int ds, de; 13 14 int nnum[1002]; 15 int nCnt; 16 17 int m,n; 18 19 int getV(char a) { 20 if(a >= ‘0‘ && a <= ‘9‘) { 21 return a - ‘0‘; 22 } 23 else if(a >= ‘A‘ && a <= ‘Z‘) { 24 return a - ‘A‘ + 10; 25 } 26 else if(a >= ‘a‘ && a <= ‘z‘) { 27 return a - ‘a‘ + 10; 28 } 29 30 } 31 32 char putV(int a) { 33 if(a >= 0 && a <= 9) { 34 return a + ‘0‘; 35 } 36 else { 37 return a - 10 + ‘a‘; 38 } 39 } 40 41 int div() { 42 int j = 0; 43 int ci = 0; 44 for(int i = 0; i < de; i++) { 45 int ben = ci*m + tDiv[i]; 46 tDiv[j++] = ben/n; 47 ci = ben%n; 48 } 49 int i = 0; 50 while(tDiv[i] == 0) { 51 i++; 52 } 53 ds = i; 54 return ci; 55 } 56 57 58 int main(int argc, char const *argv[]) 59 { 60 while(scanf("%d %d",&m,&n) != EOF) { 61 scanf("%s",mnum); 62 int lenm = strlen(mnum); 63 64 for(int i = 0; i < lenm; i++) { 65 tDiv[i] = getV(mnum[i]); 66 } 67 68 ds = 0, de = lenm; 69 tDiv[de] = -1; 70 int j = 0; 71 memset(nnum, 0, sizeof(nnum)); 72 while(ds != de) { 73 nnum[j++] = div(); 74 } 75 for(int i = j-1; i >= 0; i--) { 76 printf("%c",putV(nnum[i])); 77 } 78 puts(""); 79 } 80 return 0; 81 }
虽然通过，但耗时70ms
考虑有没有优化的空间，观察一下div 函数，每回都从0开始除，其实每回应该从ds除就好了。50行也应该从ds开始找，但注意j初值也为ds
另外，输出也可改一下
代码如下
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <iostream> 5 6 using namespace std; 7 8 char mnum[1002]; 9 10 11 int tDiv[1002]; 12 int ds, de; 13 14 int nnum[1002]; 15 int nCnt; 16 17 char ans[1002]; 18 int m,n; 19 20 int getV(char a) { 21 if(a >= ‘0‘ && a <= ‘9‘) { 22 return a - ‘0‘; 23 } 24 else if(a >= ‘A‘ && a <= ‘Z‘) { 25 return a - ‘A‘ + 10; 26 } 27 else if(a >= ‘a‘ && a <= ‘z‘) { 28 return a - ‘a‘ + 10; 29 } 30 31 } 32 33 char putV(int a) { 34 if(a >= 0 && a <= 9) { 35 return a + ‘0‘; 36 } 37 else { 38 return a - 10 + ‘a‘; 39 } 40 } 41 42 int div() { 43 int j = ds; 44 int ci = 0; 45 for(int i = ds; i < de; i++) { 46 int ben = ci*m + tDiv[i]; 47 tDiv[j++] = ben/n; 48 ci = ben%n; 49 } 50 int i = ds; 51 while(tDiv[i] == 0) { 52 i++; 53 } 54 ds = i; 55 return ci; 56 } 57 58 59 int main(int argc, char const *argv[]) 60 { 61 while(scanf("%d %d",&m,&n) != EOF) { 62 scanf("%s",mnum); 63 int lenm = strlen(mnum); 64 65 for(int i = 0; i < lenm; i++) { 66 tDiv[i] = getV(mnum[i]); 67 } 68 69 ds = 0, de = lenm; 70 tDiv[de] = -1; 71 int j = 0; 72 memset(nnum, 0, sizeof(nnum)); 73 while(ds != de) { 74 nnum[j++] = div(); 75 } 76 int p = 0; 77 for(int i = j-1; i >= 0; i--) { 78 ans[p++] = putV(nnum[i]); 79 } 80 ans[p] = ‘\0‘; 81 puts(ans); 82 } 83 return 0; 84 }
这样，时间缩短到了30ms
但代码还有优化的空间，即数组的一位不再是一位数字，而可以是多位，这样运算次数会更少。
但比较麻烦，以后再考虑吧
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原文地址：http://www.cnblogs.com/jasonJie/p/5819843.html
