st1\:*{behavior:url(#ieooui) }
对于给出的n个询问,每次求有多少个数对(x,y),满足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函数为x和y的最大公约数。
标签:stat mst clu 公约数 ret sum while ble class
st1\:*{behavior:url(#ieooui) }
对于给出的n个询问,每次求有多少个数对(x,y),满足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函数为x和y的最大公约数。
第一行一个整数n,接下来n行每行五个整数,分别表示a、b、c、d、k
共n行,每行一个整数表示满足要求的数对(x,y)的个数
100%的数据满足:1≤n≤50000,1≤a≤b≤50000,1≤c≤d≤50000,1≤k≤50000
把询问拆成四个,就和[POI2007]ZAP一样了
#pragma GCC optimize("O2") #include <cstdio> #include <algorithm> using namespace std; char buf[10000000], *ptr = buf - 1; inline int readint(){ int n = 0; while(*++ptr < ‘0‘ || *ptr > ‘9‘); while(*ptr <= ‘9‘ && *ptr >= ‘0‘) n = (n << 1) + (n << 3) + (*ptr++ & 15); return n; } const int maxn = 50000 + 10; bool mark[maxn] = {false}; int mu[maxn], sum[maxn]; int pri[maxn], prn = 0; void shai(){ mu[1] = 1; for(int i = 2; i <= 50000; i++){ if(!mark[i]){ pri[++prn] = i; mu[i] = -1; } for(int j = 1; j <= prn && pri[j] * i <= 50000; j++){ mark[i * pri[j]] = true; if(i % pri[j] == 0){ mu[i * pri[j]] = 0; break; } else mu[i * pri[j]] = -mu[i]; } } sum[0] = 0; for(int i = 1; i <= 50000; i++) sum[i] = sum[i - 1] + mu[i]; } inline int work(int n, int m){ if(n > m) swap(n, m); int ans = 0, pos; for(int i = 1; i <= n; i = pos + 1){ pos = min(n / (n / i), m / (m / i)); ans += (sum[pos] - sum[i - 1]) * (n / i) * (m / i); } return ans; } int main(){ fread(buf, sizeof(char), sizeof(buf), stdin); shai(); int n = readint(); int a, b, c, d, k; while(n--){ a = readint(); b = readint(); c = readint(); d = readint(); k = readint(); printf("%d\n", work(b / k, d / k) + work((a - 1) / k, (c - 1) / k) - work((a - 1) / k, d / k) - work(b / k, (c - 1) / k)); } return 0; }
标签:stat mst clu 公约数 ret sum while ble class
原文地址:http://www.cnblogs.com/ruoruoruo/p/7630106.html