| Run ID | 作者 | 问题 | 语言 | 测评结果 | Time | Memory | 代码长度 | 提交时间 |
|---|---|---|---|---|---|---|---|---|
| 2632 | 孙浚轩 | 进制转换2 | C++ | Wrong Answer | 0 MS | 272 KB | 845 | 2024-01-20 13:07:16 |
Tests(0/1):
Code:
#include<iostream> #include<cstdio> #include<string> #include<cmath> using namespace std; int ctoi(char c){ if(c>='0'&&c<='9') return c-'0'; else return c-'A'+10; } char itoc(int n){ if(n>=0&&n<=9) return '0'+n; else return 'A'+n-10; } int mtoten(string num,int m){ int d=0; double p=0; for(int i=num.length()-1;i>=0;i--){ d+=ctoi(num[i])*pow(m,p); p++; } return d; } string tenton(int d,int n){ string res=""; while(d>0){ int rem=d%n; res=itoc(rem)+res; d/=n; } return res; } int main(){ int m,n; string x; while(scanf("%d",&n)!=EOF){ cin>>n>>x; int dec=mtoten(x,m); string res=tenton(dec,n); cout<<res<<endl; } return 0; }
Run Info:
------Input------
2 10 11111111 3 16 1212120 4 8 1230123 5 11 43210 6 12 342150 7 13 1234560 8 14 12345670 9 15 123456780 10 2 11 11 17 12345A 12 18 23456AB 13 19 12345678ABC 14 20 23456ABCD 15 21 4567ABDE 16 22 12345DEF 17 23 2314EFG 18 24 12345ABH 19 25 456CDEI 20 26 13567ABDJ 21 27 32145CBADK 22 28 923456781CDBAEFGHIKJL 23 29 54321ABM 24 30 234567CDFGN 25 31 ABCDEO 26 32 ABCDP 27 33 BGDFAQ 28 34 34AFGRDE 29 35 678SAKB 30 36 TEGH12 31 3 345GHJU 32 4 VASDBC 33 5 FGC4W12H 34 6 ABCX34DFR 35 7 6H8Y34GRW 36 9 XYZZA7
------Answer-----
255 555 15433 2224 14956 57b64 514320 4bb27b6 1011 25b54 3affd5 9a00b5ehf 2bbaaf81 8ea972e 2f5hah9 842ef6 3e4i72n kfgi45 3iacap1p 8j6a3da55 219m58bk1b9n8cqg5d47 10io49cq 6q1or35ten 3hiqj1 4hjnj 48g3sk rjb8vng 20s7fok bukt5w 21011220002022001011 332223203111230 41310012401323330 103130240520530445 3036240551316664 10374545654665
------Your output-----
1 3 1 A B C D E F G H I J K L M N O P Q E B 2 1010 30 32 43 44 5