[hdu4710 Balls Rearrangement]分段统计

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; }