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传送门题中有XA 和 X+B 两种增长方法当XA < X+B 而且 XA < Y时，使用第一种方法否则用第二种方法，第二种方法可以直接计算出答案，不用累加#include<stdio.h>#include<iostream>#include<cstring>#define ll long longusing namespace std;int main(){ ll x,y; ll a,b; cin>>x

AtCoder Beginner Contest 180传送门A - box#include<iostream>using namespace std;int main(){ int n,a,b; cin>>n>>a>>b; cout<<n-a+b<<endl; return 0;}B - Various distances解题思路：按题目中的公式进行求解，注意The Eucli

Sample Input 12 37 9Sample Output 14Sample Input 23 03 4 5Sample Output 25Sample Input 310 10158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202Sample Output 3 292638192解题思路：求<=k的切割次数中.

解题思路：Sample Input 1101Sample Output 14None of 7, 77, and 777 is a multiple of 101, but 7777 is.对这个例子分析先从7尝试对101取余，然后7*10+7 = 77，77对101取余后 乘10 +7 = 777 ，一直到 777 对101取余后，777变为70，70又继续 乘10+7 = 707，在这里对101取余为 0 ，符合 答案为 4 。我们可以得出规律，k的范围是[1, 1e6]，假如该数能被.

C.UbiquitySample Input 12Sample Output 12Sample Input 21Sample Output 20Sample Input 3869121Sample Output 32511445题解：输入一个N，0<=Ai<=9，所以一共10N 种情况，序列中元素个数为N，序列中一定存在 0 和 9，要得到至少有一个0和一个9的所有情况，思路使用总共的情况减去只有一个 0 、只有 一个 9 、或者 0 和 9 都没有的情况。用 10

Sample Input 16 2 4Sample Output 12Sample Input 27 4 3Sample Output 21Takahashi is now at coordinate 7. It is optimal to make, for example, the following moves:Move from coordinate 7 to 4Move from coordinate 4 to 7Move from coordinate 7 to 4Mo.

Sample Input 1Copy4 41 14 4..#...#..#...#..Sample Output 1Note that he cannot walk diagonally.Sample Input 2Copy4 41 44 1.##.########.##.Sample Output 2Copy-1He cannot move from there.Sample Input 34 42 23 3…………Sample ..

Sample Input 1cabaccabcSample Output 11Sample Input 2codeforcesatcoderSample Output 26#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<string>#includ..

c-Count OrderEditorialTime Limit: 2 sec/ Memory Limit:1024MBScore: 300 pointsProblem StatementWe have two permutations Pand Q of size N(that is, Pand Q are both rearrangements of(1，2，…，N)).There are N！possible permutatons of size N.Among them,let Pan