8.26 T2 日记和欧拉函数（欧拉函数）

本文主要是介绍8.26 T2 日记和欧拉函数（欧拉函数），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

http://cplusoj.com/d/senior/p/NOD2301B

发现 $x\le B$ 时答案是 $x$

$x > B + 500$ 左右答案是1

我们预处理中间的就行

预处理直接暴力做，求 $\max \phi$ 的话相当于求小于它的质数

#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL#define debug(...) fprintf(stdout, ##__VA_ARGS__)#define debag(...) fprintf(stderr, ##__VA_ARGS__)
#else#define debug(...) void(0)#define debag(...) void(0)
#endif
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
#define fi first
#define se second
//#define M
//#define mo
#define N 510
int n, m, i, j, k, T, B;
int mx, phi[N], f[N], sm[N], l, r;
int ans;  
map<int, int>mp; int qry(int x) {if(mp[x]) return mp[x]; int ans = 1, i, j; 
//	debug("phi[%lld] = ", x); for(i = 2; i * i <= x; ++i) if(x % i == 0) {j = 1; while(x % i == 0) x /= i, j *= i; j /= i; j *= (i - 1); ans *= j; }if(x > 1) ans *= (x - 1); 
//	debug("%lld\n", ans); return mp[x] = ans; 
}signed main()
{#ifdef LOCALfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endif
//	srand(time(NULL));
//	T=read();
//	while(T--) {
//
//	}T = read(); B = read(); for(j = 1 + B; ; --j) if(j == 1 || qry(j) == j - 1) break; 
//	debug("Out\n"); for(i = 1; i <= 400; ++i) {k = i + B; phi[i] = qry(k); if(phi[i] == k - 1) j = k; mx = j - 1 - B; f[i] = k; if(i <= 6) debug("[%lld]%lld %lld | %lld\n", k, mx, phi[i], j); if(mx <= 0) continue; while(mx-- && f[i] != 1) f[i] = qry(f[i]); if(i <= 6) debug("[%lld] %lld\n", k, f[i]); }for(i = 1; i <= 400; ++i) sm[i] = sm[i - 1] + f[i]; while(T--) {l = read(); r = read(); ans = 0; if(r <= B) { printf("%lld\n", (l + r) * (r - l + 1) / 2); continue; }if(l <= B) ans += (l + B) * (B - l + 1) / 2, l = B + 1; if(l >= B + 400) { printf("%lld\n", r - l + 1); continue; }if(r >= B + 400) ans += (r - (B + 400) + 1), r = B + 400 - 1; debug("[%lld %lld] %lld\n", l, r, ans); l -= B; r -= B; ans += sm[r] - sm[l - 1]; debug("[%lld %lld]\n", l, r); printf("%lld\n", ans); }return 0;
}