数论练习
欧拉函数:
欧拉函数筛法:
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const int N = 5e6 + 5; ull ans[N]; void count(int n) //求1~maxv的所有欧拉函数 { memset(ans, 0, sizeof(ans)); for (int i = 2; i < n; i++) { if (!ans[i]) //第一个ans[i] == 0 的都是素数 { for (int j = i; j < n; j += i) //素数的倍数 { if (!ans[j]) ans[j] = j; //对素数倍数赋值 ans[j] = ans[j] / i * (i - 1); //求欧拉函数的值 } } } for (int i = 2; i < n; i++) ans[i] = ans[i - 1] + ans[i]; }
1007:
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#include <iostream> #include <vector> #include <algorithm> #include <string> #include <set> #include <queue> #include <map> #include <sstream> #include <cstdio> #include <cstring> #include <numeric> #include <cmath> #include <iomanip> #include <deque> #include <bitset> #include <cassert> #include <unordered_set> #include <unordered_map> #define ll long long #define ull unsigned long long #define pii pair<int, int> #define pll pair<ll,ll> #define rep(i,a,b) for(int i=a;i<=b;i++) #define dec(i,a,b) for(int i=a;i>=b;i--) #define forn(i, n) for(int i = 0; i < int(n); i++) using namespace std; int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } }; const long long INF = 0x7f7f7f7f7f7f7f7f; const int inf = 0x3f3f3f3f; const double pi = acos(-1.0); const double eps = 1e-6; inline ll read() { ll x = 0; bool f = true; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); } while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar(); return f ? x : -x; } ll gcd(ll m, ll n) { return n == 0 ? m : gcd(n, m % n); } ll qpow(ll m, ll k, ll mod) { ll res = 1, t = m; while (k) { if (k & 1) res = res * t % mod; t = t * t % mod; k >>= 1; } return res; } /**********************************************************/ const ll mod = 1e9 + 7; const int N = 5e6 + 5; ull ans[N]; void count(int n) //求1~maxv的所有欧拉函数 { memset(ans, 0, sizeof(ans)); for (int i = 2; i < n; i++) { if (!ans[i]) //第一个ans[i] == 0 的都是素数 { for (int j = i; j < n; j += i) //素数的倍数 { if (!ans[j]) ans[j] = j; //对素数倍数赋值 ans[j] = ans[j] / i * (i - 1); //求欧拉函数的值 } } } for (int i = 2; i < n; i++) //求平方和 ans[i] = ans[i - 1] + ans[i] * ans[i]; } int main() { #ifdef _DEBUG freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif count(N); int T; cin >> T; rep(i,1,T) { int a, b; cin >> a >> b; printf("Case %d: %llu\n", i, ans[b] - ans[a - 1]); } }