I - Olympiad
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了I - Olympiad相关的知识,希望对你有一定的参考价值。
InputThe first line of the input is a single integer T (T≤1000)T (T≤1000), indicating the number of testcases.
For each test case, there are two numbers aa and bb, as described in the statement. It is guaranteed that 1≤a≤b≤1000001≤a≤b≤100000.
OutputFor each testcase, print one line indicating the answer.
Sample Input
2 1 10 1 1000
Sample Output
10 738
打表大法好呀
1 #include <iostream> 2 using namespace std; 3 #include<string.h> 4 #include<set> 5 #include<stdio.h> 6 #include<math.h> 7 #include<queue> 8 #include<map> 9 #include<algorithm> 10 #include<cstdio> 11 #include<cmath> 12 #include<cstring> 13 #include <cstdio> 14 #include <cstdlib> 15 #include<cstring> 16 int a[1000010],b[15]; 17 int guanyinzuolian(int n) 18 { 19 memset(b,0,sizeof(b)); 20 while(n!=0) 21 { 22 if(b[n%10]) 23 return 0; 24 b[n%10]++; 25 n/=10; 26 } 27 return 1; 28 } 29 void lubenwei() 30 { 31 int sum=0; 32 for(int i=1;i<=100000;i++) 33 a[i]=a[i-1]+guanyinzuolian(i); 34 } 35 int main() 36 { 37 memset(a,0,sizeof(a)); 38 lubenwei(); 39 int t; 40 cin>>t; 41 while(t--) 42 { 43 int n,m; 44 cin>>n>>m; 45 cout<<a[m]-a[n-1]<<endl; 46 } 47 }
以上是关于I - Olympiad的主要内容,如果未能解决你的问题,请参考以下文章
Codeforces Round #680 (Div. 2, based on Moscow Team Olympiad)ABCD
Codeforces Round #626 (Div. 1, based on Moscow Open Olympiad in Informatics)B(位运算,二分查找)
Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) A. Fraction
Codeforces Round #626 (Div. 2, based on Moscow Open Olympiad in Informatics)(A-C)