• C. Yet Another Counting Problem


    You are given two integers aa and bb, and qq queries. The ii-th query consists of two numbers lili and riri, and the answer to it is the number of integers xx such that lixrili≤x≤ri, and ((xmoda)modb)((xmodb)moda)((xmoda)modb)≠((xmodb)moda). Calculate the answer for each query.

    Recall that ymodzymodz is the remainder of the division of yy by zz. For example, 5mod3=25mod3=2, 7mod8=77mod8=7, 9mod4=19mod4=1, 9mod9=09mod9=0.

    Input

    The first line contains one integer tt (1t1001≤t≤100) — the number of test cases. Then the test cases follow.

    The first line of each test case contains three integers aa, bb and qq (1a,b2001≤a,b≤200; 1q5001≤q≤500).

    Then qq lines follow, each containing two integers lili and riri (1liri10181≤li≤ri≤1018) for the corresponding query.

    Output

    For each test case, print qq integers — the answers to the queries of this test case in the order they appear.

    Example
    input
    Copy
    2
    4 6 5
    1 1
    1 3
    1 5
    1 7
    1 9
    7 10 2
    7 8
    100 200
    
    output
    Copy
    0 0 0 2 4 
    0 91 
    

     在这个问题里面,(r-l+1)/LCM和r/LCM-(l-1)/LCM差别是很大的。。。

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <iomanip>
    #include <deque>
    #include <bitset>
    //#include <unordered_set>
    //#include <unordered_map>
    //#include <bits/stdc++.h>
    //#include <xfunctional>
    #define ll              long long
    #define PII             pair<int, int>
    #define rep(i,a,b)      for(int  i=a;i<=b;i++)
    #define dec(i,a,b)      for(int  i=a;i>=b;i--)
    #define pb              push_back
    #define mk              make_pair
    using namespace std;
    int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = 3.14159265358979;
    const int mod = 998244353;
    const int N = 1e6+5;
    //if(x<0 || x>=r || y<0 || y>=c)
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    ll lcm(ll m, ll n)
    {
        return m*n / gcd(m, n);
    }
    int main()
    {
        int T;
        cin >> T;
        while (T--)
        {
            ll a, b, q;
            cin >> a >> b >> q;
            ll LCM = lcm(a, b);
            vector<ll> v(LCM+1,0);
            for (int i = 1; i <= LCM; i++)
            {
                if ((i%a)%b != (i%b)%a)
                {
                    v[i] = v[i - 1] + 1;
                }
                else
                    v[i] = v[i - 1];
            }
    
            while (q--)
            {
                ll l, r;
                l = read(), r = read();
                ll t1 = r/LCM*v[LCM]+v[r%LCM];
                ll t2 = (l - 1) / LCM*v[LCM] + v[(l - 1) % LCM];
                cout << t1 - t2<< " ";
            }
            cout << endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dealer/p/12839781.html
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