1 solutions
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0
C++ :
#include <iostream> using namespace std; int main() { int n,ge,shi,bai; cin>>n; ge=n%10;//个位 shi=(n/10)%10;//十位 bai=n/100;//百位 //要保留前导0就不能用这个方法 // n=ge*100+shi*10+bai; // cout<<n<<endl; //保留前导0输出的正确姿势 cout<<ge<<shi<<bai<<endl; return 0; }Pascal :
var a,a1,a2,a3:integer; begin read(a); a1:=a div 100; a2:=a div 10 mod 10; a3:=a mod 10; write(a3,a2,a1); end.
- 1
Information
- ID
- 17044
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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