2 solutions
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1
P1980 C++
#include<bits/stdc++.h> using namespace std; int n1,n2; int cnt=0; int cot(int n,int m){ int cntr=0; while(n!=0){ if(n%10==m){ cntr++; } n/=10; } return cntr; } int main(){ cin>>n1>>n2; for(int i=1;i<=n1;i++){ cnt+=cot(i,n2); } cout<<cnt; return /*164*/0; }
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1
#include<bits/stdc++.h> #define Goose main using namespace std; #define qUark int #define g ; #define QUARK return #define Quark cin #define quarK cout #define qq >> #define kk << #define quArk for #define quaRk while #define QUark if #define ss { #define ee } #define aa ( #define rr ) qUark n,x,ans=0 g qUark Goose aa rr ss Quark qq n qq x g quArk aa qUark i=1 g i<=n g i++ rr ss qUark k=i g quaRk aa k rr ss QUark aa k%10==x rr ans++ g k/=10 g ee ee quarK kk ans kk "\n" g QUARK 0 g ee
Goose语言精选代码(保a)
- 1
Information
- ID
- 1131
- Time
- 1000ms
- Memory
- 125MiB
- Difficulty
- 3
- Tags
- # Submissions
- 25
- Accepted
- 19
- Uploaded By