两个n(2~35)进制的求和输出用的和用n进制表示

#include<stdio.h> #include<string.h> #include <stdlib.h> int zhan(char t) { int a; char s[3]=""; int b; if(isdigit(t)) { b=t-'0'; a=0+b; } else { b=t-'a'; a=10+b; } return a; } char zhuan(int t) { char c; if(t>=0&&t<=9) { c=t+'0'; } else { c='a'+t-10; } return c; } int main() { int n; char ch; int b; int c; int flag=0; int len1; int len2; int len3; int i; int j1; int j2; char s[1000]; char s1[1000]; char s2[1000]; char s3[1000]=""; int d[1000]={0}; scanf("%d",&n); scanf("%s",s); scanf("%s",s1); len1=strlen(s); len2=strlen(s1); if(len1>len2) { len3=len1; len1=len2; len2=len3; strcpy(s2,s); strcpy(s,s1); strcpy(s1,s2); } for(i=0;i<len1;i++) { b=zhan(s[len1-1-i]); c= zhan(s1[len2-1-i]); d[i]=(flag+(b+c))%n; flag=(flag+(b+c))/n; } for(i=len1;i<len2;i++) { b=zhan(s1[len2-1-i]); c=0; d[i]=(flag+(b+c))%n; flag=(flag+(b+c))/n; } if(flag!=0) { d[i]=flag; for(i=0;i<len2+1;i++) { s3[i]=zhuan(d[len2+1-1-i]); } } else { for(i=0;i<len2;i++) { s3[i]=zhuan(d[len2-1-i]); } } printf("%s",s3); return 0; }