Codeforce 25D - HackMD

--- title: Codeforce 25D description: "Codeforce 25D math number-theory" ---  # Codeforce 1342C Yet Another Counting Problem Today we are gonna take a look of problem Codeforce 1342C [Problem Link](https://codeforces.com/problemset/problem/1342/C) > **Problem** You are given two integers a and b, and q queries. The i-th query consists of two numbers li and ri, and the answer to it is the number of integers x such that li≤x≤ri, and $((x\mod a)\mod b)\neq((x\mod b)\mod a)$. Calculate the answer for each query. Recall that ymodz is the remainder of the division of y by z. For example, 5mod3=2, 7mod8=7, 9mod4=1, 9mod9=0. **Input** The first line contains one integer t (1≤t≤100) — the number of test cases. Then the test cases follow. The first line of each test case contains three integers a, b and q (1≤a,b≤200; 1≤q≤500). Then q lines follow, each containing two integers li and ri (1≤li≤ri≤$10^{18}$) for the corresponding query. **Output** For each test case, print q integers — the answers to the queries of this test case in the order they appear. ### Idea The queried range is too big($10^{18}$), so we can't check each number directly. There must be some math in it. Observe that $\forall x\in\{\mathbb{N}\cup0\}$, $((x\mod a)\mod b)\neq((x\mod b)\mod a)\iff\\(((x\mod ab)\mod a)\mod b)\neq(((x\mod ab)\mod b)\mod a)$ $i.e.$ replace $x$ with $x\mod ab$. And $ab\le40000$, so it's possible to check all $x$ where $x\le40000$ if it satisfy the condition. As the range is composed of some range of length $ab$ and a last segment with length$<ab$. Maintain a prefix sum array for $x$ in range $[0,ab-1]$, we can easily calculate the answer. (Note that I actually maintain prefix of $[0,2ab-1]$ since I need to calculate the number of last segment). ### Code: ```c++ #pragma GCC optimize(1) #pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize("Ofast") #pragma GCC optimize("inline") #pragma GCC optimize("-fgcse") #pragma GCC optimize("-fgcse-lm") #pragma GCC optimize("-fipa-sra") #pragma GCC optimize("-ftree-pre") #pragma GCC optimize("-ftree-vrp") #pragma GCC optimize("-fpeephole2") #pragma GCC optimize("-ffast-math") #pragma GCC optimize("-fsched-spec") #pragma GCC optimize("unroll-loops") #pragma GCC optimize("-falign-jumps") #pragma GCC optimize("-falign-loops") #pragma GCC optimize("-falign-labels") #pragma GCC optimize("-fdevirtualize") #pragma GCC optimize("-fcaller-saves") #pragma GCC optimize("-fcrossjumping") #pragma GCC optimize("-fthread-jumps") #pragma GCC optimize("-funroll-loops") #pragma GCC optimize("-freorder-blocks") #pragma GCC optimize("-fschedule-insns") #pragma GCC optimize("inline-functions") #pragma GCC optimize("-ftree-tail-merge") #pragma GCC optimize("-fschedule-insns2") #pragma GCC optimize("-fstrict-aliasing") #pragma GCC optimize("-falign-functions") #pragma GCC optimize("-fcse-follow-jumps") #pragma GCC optimize("-fsched-interblock") #pragma GCC optimize("-fpartial-inlining") #pragma GCC optimize("no-stack-protector") #pragma GCC optimize("-freorder-functions") #pragma GCC optimize("-findirect-inlining") #pragma GCC optimize("-fhoist-adjacent-loads") #pragma GCC optimize("-frerun-cse-after-loop") #pragma GCC optimize("inline-small-functions") #pragma GCC optimize("-finline-small-functions") #pragma GCC optimize("-ftree-switch-conversion") #pragma GCC optimize("-foptimize-sibling-calls") #pragma GCC optimize("-fexpensive-optimizations") #pragma GCC optimize("inline-functions-called-once") #pragma GCC optimize("-fdelete-null-pointer-checks") #include <bits/stdc++.h> using namespace std; #define rep(i,a,n) for(int i=a;i<n;i++) #define per(i,a,n) for(int i=n-1;i>=a;i--) #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second #define SZ(x) ((int)(x).size()) #define min(a,b) (((a)<(b))?(a):(b)) #define max(a,b) (((a)>(b))?(a):(b)) typedef vector<int> VI; typedef long long ll; typedef pair<int,int> PII; typedef double db; mt19937 mrand(random_device{}()); const ll mod=1000000007; int rnd(int x){return mrand()%x;} ll powmod(ll a,ll b){ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll gcd(ll a, ll b){return b?gcd(b,a%b):a;} #define rank oiajgpowsdjg const int N = 100; int parent[N], rank[N]; inline void dsinit(int n) {for (int i = 0; i < n; i++)parent[i] = i;memset(rank, 0, sizeof rank);} inline int dsfind(int e) {return parent[e] == e ? e : parent[e] = dsfind(parent[e]);} inline void dsunion(int s1, int s2) {if (rank[s1] < rank[s2])swap(s1, s2);parent[s2] = s1;if (rank[s1] == rank[s2]) rank[s1]++;} //head int t,a,b,q,pre[80050]; ll l,r; main(void) { cin.tie(0); ios_base::sync_with_stdio(0); cin>>t;while(t--){ cin>>a>>b>>q; pre[0]=0;rep(i,1,2*a*b+1)pre[i]=pre[i-1]+(i%a%b!=i%b%a?1:0); while(q--){ cin>>l>>r;ll ans=0; ans+=1ll*(r-l+1)/(1ll*(a*b))*1ll*pre[a*b-1]; ans+=pre[(l-1)%(1ll*a*b)+(r-l+1)%(1ll*a*b)]-pre[(l-1)%(1ll*a*b)]; cout<<ans<<' '; } cout<<'\n'; } return 0; } ``` [Submission](https://codeforces.com/contest/1342/submission/87735792)