题意：
求【a，b】间的欧拉函数和。这道题卡内存，只能开一个数组。
思路：
ϕ(n) = n * （p-1)/p * ... 可利用线性筛法求出所有ϕ(n) 。
#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <iomanip>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <cctype>
#include <queue>
#include <cmath>
#include <list>
#include <map>
#include <set>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000") //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue typedef long long ll;
typedef unsigned long long ull; typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行
#define REP(i , j , k) for(int i = j ; i < k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //
const ll nmos = 0x80000000LL; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //
const double PI=acos(-1.0); template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
}
// #define _DEBUG; //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/
const int maxn = ; ll sum[maxn];
/*
const int maxn = 3000009;
int prime[316819];
int phi[maxn];
ll sum[maxn];
bool flag[maxn];
void getphi(){
phi[1] = 1,sum[1] = 1;
int tot = 0;
for(int i=2; i<maxn; i++){
if(flag[i] == 0){
prime[++tot] = i;
phi[i] = i-1;
}
for(int j=1; i * prime[j] < maxn && j<=tot; j++){
flag[i*prime[j]] = 1;
if(i%prime[j] == 0){
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
else phi[i * prime[j]] = phi[i]*(prime[j] - 1);
}
sum[i] = sum[i-1] + phi[i];
} }
*/
void getphi(){
sum[] = ;
for(int i=; i<maxn; i++){
if(!sum[i]){
for(int j=i; j<maxn; j+=i){
if(!sum[j])sum[j] = j;
sum[j] = sum[j] / i * (i-);
}
}
}
for(int i=; i<maxn; i++)sum[i] += sum[i-];
}
int main(){
int a,b;
getphi();
while(~scanf("%d%d", &a, &b)){
printf("%lld\n", sum[b] - sum[a-]);
} return ;
}
HDU - 2824
还有POJ的2478也是区间累计欧拉函数的和
#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <iomanip>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <cctype>
#include <queue>
#include <cmath>
#include <list>
#include <map>
#include <set>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000") //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue typedef long long ll;
typedef unsigned long long ull; typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行
#define REP(i , j , k) for(int i = j ; i < k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //
const ll nmos = 0x80000000LL; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //
const double PI=acos(-1.0); template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
}
// #define _DEBUG; //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/
const int maxn = ; ll sum[maxn];
void getphi(){
sum[] = ;
for(int i=; i<maxn; i++){
if(!sum[i]){
for(int j=i; j<maxn; j+=i){
if(!sum[j])sum[j] = j;
sum[j] = sum[j] / i * (i-);
}
}
}
for(int i=; i<maxn; i++)sum[i] += sum[i-];
}
int main(){
int a;
getphi();
while(~scanf("%d", &a) && a){
printf("%lld\n", sum[a]);
} return ;
}
POJ2478