进制转换
将n 进制数x，转换为m 进制数y输出.(n,m<=20)
输入格式
x<n>m
输出格式
x<n>=y<m>
样例输入
48<10>8
样例输出
48<10>=60<8>
没啥好讲的，先转化成10进制，再转化成其他进制。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 using namespace std; 7 #define rep(i, a, n) for(int i = a; i <= n; ++i) 8 #define per(i, n, a) for(int i = n; i >= a; --i) 9 typedef long long ll; 10 const int maxn = 1e2 + 5; 11 char ax[20] ={'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'}; 12 char a[maxn], num[maxn], nnum[maxn]; 13 ll pre, now, chanum; 14 void retu() //先转换成10进制 15 { 16 ll x = 1; 17 per(i, strlen(num) - 1, 0) 18 { 19 int y = num[i] >= 'A' ? num[i] - 'A' + 10 : num[i] - '0'; 20 chanum += y * x; 21 x *= pre; 22 } 23 } 24 void change(ll num) //在转换成n进制 25 { 26 int pos = 0; 27 while(num > 0) 28 { 29 int x = num % now; 30 nnum[++pos] = x > 9 ? ax[x - 10] : x + '0'; 31 num /= now; 32 } 33 } 34 int main() 35 { 36 scanf("%s", a); 37 int i = 0; 38 while(a[i] != '<') {num[i] = a[i]; i++;} 39 ++i; 40 while(a[i] != '>') {pre = pre * 10 + a[i] - '0'; i++;} 41 ++i; 42 while(i < strlen(a)) {now = now * 10 + a[i] - '0'; i++;} 43 retu(); 44 change(chanum); 45 printf("%s<%lld>=", num, pre); 46 per(i, strlen(nnum + 1), 1) printf("%c", nnum[i]); 47 printf("<%lld>\n", now); 48 return 0; 49 }
