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 a n s = l ∗ 1 0 a + ( l − 1 ) ∗ 1 0 b + . . . . . + r ∗ 1 0 0 ans=l*10^a+(l-1)*10^b+…..+r*10^0 ans=l∗10a+(l−1)∗10b+.....+r∗100, 10 % 1 = 9 10\%1=9 10%1=9因此变为 a n s = l + l − 1 + l − 2 + . . . + r ans=l+l-1+l-2+…+r ans=l+l−1+l−2+...+r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;
}

本文地址：https://blog.csdn.net/kosf_/article/details/108151767