[逆元] luogu 4942
程序员文章站
2023-12-29 16:55:10
题目题目链接:https://www.luogu.com.cn/problem/P4942思路这道题最后的ans的形式为ans=l∗10a+(l−1)∗10b+.....+r∗100ans=l*10^a+(l-1)*10^b+.....+r*10^0ans=l∗10a+(l−1)∗10b+.....+r∗100,10%1=910\%1=910%1=9因此变为ans=l+l−1+l−2+...+rans=l+l-1+l-2+...+rans=l+l−1+l−2+...+r用等差数列即可代码#inc...
题目
题目链接:https://www.luogu.com.cn/problem/P4942
思路
这道题最后的ans的形式为,因此变为用等差数列即可
代码
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<ctime>
#include<iostream>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<iomanip>
#include<list>
#include<bitset>
#include<sstream>
#include<fstream>
#include<complex>
#include<algorithm>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define int long long
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 9;
signed main(){
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int t;
cin>>t;
while(t--){
int l,r;
cin>>l>>r;
cout<<((l%mod*(r-l+1)%mod)%mod+((r-l+1)%mod*(r-l+mod)%mod)%mod*5%mod)%mod<<endl;
}
return 0;
}
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