[POI2007]ZAP-Queries 数学
题目描述
Byteasar the Cryptographer works on breaking the code of BSA (Byteotian Security Agency). He has alreadyfound out that whilst deciphering a message he will have to answer multiple queries of the form"for givenintegers aaa, bbb and ddd, find the number of integer pairs (x,y)(x,y)(x,y) satisfying the following conditions:
1≤x≤a1\le x\le a1≤x≤a,1≤y≤b1\le y\le b1≤y≤b,gcd(x,y)=dgcd(x,y)=dgcd(x,y)=d, where gcd(x,y)gcd(x,y)gcd(x,y) is the greatest common divisor of xxx and yyy".
Byteasar would like to automate his work, so he has asked for your help.
TaskWrite a programme which:
reads from the standard input a list of queries, which the Byteasar has to give answer to, calculates answers to the queries, writes the outcome to the standard output.
FGD正在破解一段密码,他需要回答很多类似的问题:对于给定的整数a,b和d,有多少正整数对x,y,满足x<=a,y<=b,并且gcd(x,y)=d。作为FGD的同学,FGD希望得到你的帮助。
输入输出格式
输入格式:The first line of the standard input contains one integer nnn (1≤n≤50 0001\le n\le 50\ 0001≤n≤50 000),denoting the number of queries.
The following nnn lines contain three integers each: aaa, bbb and ddd(1≤d≤a,b≤50 0001\le d\le a,b\le 50\ 0001≤d≤a,b≤50 000), separated by single spaces.
Each triplet denotes a single query.
输出格式:Your programme should write nnn lines to the standard output. The iii'th line should contain a single integer: theanswer to the iii'th query from the standard input.
输入输出样例
2 4 5 2 6 4 3
3 2
![](https://img2018.cnblogs.com/blog/1473307/201902/1473307-20190213094749326-1788531776.png)
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<time.h> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 200005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) #define mclr(x,a) memset((x),a,sizeof(x)) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 98765431; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-5 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } int sqr(int x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ int n; int mu[maxn+10], vis[maxn+10]; int sum[maxn + 10]; void init() { for (int i = 1; i < maxn; i++)mu[i] = 1, vis[i] = 0; for (int i = 2; i < maxn; i++) { if (vis[i])continue; mu[i] = -1; for (int j = 2 * i; j < maxn; j += i) { vis[j] = 1; if ((j / i) % i == 0)mu[j] = 0; else mu[j] *= -1; } } for (int i = 1; i < maxn; i++)sum[i] = sum[i - 1] + mu[i]; } int main() { // ios::sync_with_stdio(0); init(); n = rd(); while (n--) { int a = rd(), b = rd(), d = rd(); ll ans = 0; for (int l = 1, r; l <= (min(a, b) / d); l = r + 1) { r = min((a / d) / (a / d / l), (b / d) / (b / d / l)); ans += 1ll * (sum[r] - sum[l - 1])*(a / d / l)*(b / d / l); } cout << (ll)ans << endl; } return 0; }