题意：求∑|i%a-i%b|，0≤i<n
思路：复杂度分析比较重要，不细想还真不知道这样一段段跳还真的挺快的=.=
- 令p=lcm(a,b)，那么p就是|i%a-i%b|的循环节。考虑计算n的答案，令答案为f(n)，则最后的结果可以表示为n/p*f(p)+f(n%p)
- 考虑计算f(n)，设两个指针i、j，表示当前累加答案的起始位置分别对a和b取模的结果，则每次i与j中间的一个可以变成0，跳过的一段区间中答案是相同的。
- 复杂度分析，令a<b，则每次跳过的区间长度近似等于a，则总的复杂度为O(n/a)=O(lcm(a,b)/a)=O(b)，也就是说复杂度大致为O(max(a,b))
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#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define X first
#define Y second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define copy(a, b) memcpy(a, b, sizeof(a))
typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-12;
/* -------------------------------------------------------------------------------- */
ll a, b, n;
ll gcd(ll a, ll b) {
return b? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
return a * b / gcd(a, b);
}
ll solve(ll n) {
ll p = 0, q = 0, c = 0, ans = 0;
while (c < n) {
ll buf = min(a - p, b - q);
umin(buf, n - c);
c += buf;
ans += buf * abs(p - q);
p = (p + buf) % a;
q = (q + buf) % b;
}
return ans;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
int T;
cin >> T;
while (T --) {
cin >> n >> a >> b;
ll p = lcm(a, b);
cout << (p <= n? n / p * solve(p) : 0) + solve(n % p) << endl;
}
return 0;
}
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时间： 2024-08-05 15:13:33