A.水题 数字翻转，将每一位大于等于5的数字t翻转成9-t，注意不要有前导0，且翻转后数字的位数不变（即9999->9000...刚开始以为应该翻转成0了= =）
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 using namespace std; 6 int main() 7 { 8 char x[20]; 9 cin >> x; 10 int len = strlen(x); 11 for(int i = 0; i < len; i++) 12 if(x[i] >= '5') x[i] = '9' - x[i] + '0'; 13 if(x[0] == '0') x[0] = '9'; 14 cout << x << endl; 15 return 0; 16 } 17 #include<iostream> 18 #include<cstdio> 19 #include<cstdlib> 20 #include<cstring> 21 using namespace std; 22 int main() 23 { 24 char x[20]; 25 cin >> x; 26 int len = strlen(x); 27 for(int i = 0; i < len; i++) 28 if(x[i] >= '5') x[i] = '9' - x[i] + '0'; 29 if(x[0] == '0') x[0] = '9'; 30 cout << x << endl; 31 return 0; 32 }