本题只需要先将$x$进制数转换成$10$进制数，然后再从十进制数转换成$y$进制数就可以了。 注意C/C++提交需要高精度 cpp代码 ```cpp #include <cstdio> #include <vector> #include <queue> #include <cstring> #include <cmath> #include <map> #include <set> #include <string> #include <iostream> #include <algorithm> #include <iomanip> using namespace std; #define sd(n) scanf("%d",&n) #define sdd(n,m) scanf("%d%d",&n,&m) #define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k) #define pd(n) printf("%d\n", n) #define pc(n) printf("%c", n) #define pdd(n,m) printf("%d %d", n, m) #define pld(n) printf("%lld\n", n) #define pldd(n,m) printf("%lld %lld\n", n, m) #define sld(n) scanf("%lld",&n) #define sldd(n,m) scanf("%lld%lld",&n,&m) #define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k) #define sf(n) scanf("%lf",&n) #define sc(n) scanf("%c",&n) #define sff(n,m) scanf("%lf%lf",&n,&m) #define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k) #define ss(str) scanf("%s",str) #define rep(i,a,n) for(int i=a;i<=n;i++) #define per(i,a,n) for(int i=n;i>=a;i--) #define mem(a,n) memset(a, n, sizeof(a)) #define debug(x) cout << #x << ": " << x << endl #define pb push_back #define all(x) (x).begin(),(x).end() #define fi first #define se second #define mod(x) ((x)%MOD) #define gcd(a,b) __gcd(a,b) #define lowbit(x) (x&-x) typedef pair<int,int> PII; typedef long long ll; typedef unsigned long long ull; typedef long double ld; const int MOD = 1e9 + 7; const double eps = 1e-9; const ll INF = 0x3f3f3f3f3f3f3f3fll; const int inf = 0x3f3f3f3f; inline int read() { int ret = 0, sgn = 1; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') sgn = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { ret = ret*10 + ch - '0'; ch = getchar(); } return ret*sgn; } inline void Out(int a) //ê?3?ía1ò { if(a>9) Out(a/10); putchar(a%10+'0'); } int qpow(int m, int k, int mod) { int res = 1, t = m; while (k) { if (k&1) res = res * t % mod; t = t * t % mod; k >>= 1; } return res; } ll gcd(ll a,ll b) { return b==0?a : gcd(b,a%b); } ll lcm(ll a,ll b) { return a*b/gcd(a,b); } const int maxn = 1000; int t[maxn], A[maxn]; char str1[maxn], str2[maxn]; int n, m; void solve() { int i, len, k; len = strlen(str1); for(i=len; i>=0; --i) t[len-1-i] = str1[i] -(str1[i]<58 ? 48: str1[i]<97 ? 55: 61); for(k=0; len;) { for(i=len; i>=1; --i) { t[i-1] +=t[i]%m*n; t[i] /= m; } A[k++] = t[0] % m; t[0] /=m; while(len>0&&!t[len-1]) len--; } str2[k] =NULL; for(i=0; i<k; i++) str2[k-1-i] = A[i]+(A[i]<10 ? 48: A[i]<36 ? 55:61); } int main() { scanf("%d%d%s",&n, &m, str1); solve(); printf("%s\n",str2); return 0; } ``` Java代码 ```java import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub String z; int x,y; Scanner input=new Scanner(System.in); x=input.nextInt(); y=input.nextInt(); z = input.next(); if(z.equals("0")) { System.out.println(0); return; } BigInteger big1=Toten(z, x); String res=reverse(ToY(big1, y)); System.out.println(res); } static String reverse(String str) { String s=""; int len = str.length(); for(int i = len-1;i >= 0;i--) { s+=str.charAt(i); } return s; } static String ToY(BigInteger a,int y) { String res=""; BigInteger zero=new BigInteger("0"); BigInteger ynum=new BigInteger(""+y); int tnum; while(!a.equals(zero)) { tnum=Integer.valueOf(a.remainder(ynum).toString()); res+=inttochar(tnum); a=a.divide(ynum); } return res; } static BigInteger Toten(String str,int x) { BigInteger res=new BigInteger("0"); BigInteger tb; BigInteger tBigInteger=BigInteger.ONE; BigInteger k = BigInteger.valueOf(x); int len = str.length(); for(int i = len-1; i >= 0;i--){ tb=new BigInteger(tBigInteger.toString()); res=res.add(chartoint(str.charAt(i)).multiply(tb)); tBigInteger = tBigInteger.multiply(k); } return res; } static BigInteger chartoint(Character ch) { int t=0; BigInteger bigInteger; if(ch>='0'&&ch<='9') { t = ch-'0'; } else if(ch>='A'&&ch<='Z') { t = ch-'A'+10; } else if(ch>='a'&&ch<='z') { t = ch-'a'+36; } bigInteger=new BigInteger(""+t); return bigInteger; } static Character inttochar(int n) { if(n>=0&&n<=9) return (char) ('0'+n); else if(n>=10&&n<=35) return (char) ('A'+n-10); else return (char) ('a'+n-36); } } ```