【容斥原理+欧拉函数】 HDOJ 1695 GCD

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先转化成求区间1-b/k和区间1-d/k的gcd值为1的对数，然后用一下欧拉函数和容斥就行了。。。

#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 40005#define eps 1e-10#define mod 1000000007#define INF 999999999#define lowbit(x) (x&(-x))#define mp mark_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R  typedef long long LL;//typedef int LL;using namespace std;LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// headint phi[maxn];int p[maxn];int res, a, b;void init(void){phi[1] = 1;for(int i = 2; i <= 100000; i++) if(!phi[i])for(int j = i; j <= 100000; j += i) {if(!phi[j]) phi[j] = j;phi[j] = phi[j] / i *(i-1);}}void dfs(int pos, int dep, int mul, int flag){if(pos == dep) {res += a/mul*flag;return;}dfs(pos+1, dep, mul, flag);dfs(pos+1, dep, mul*p[pos], -flag);}void work(int __){if(a > b) swap(a, b);int cnt, n;LL ans = 0;for(int i = 1; i <= a; i++) ans += phi[i];for(int i = a+1; i <= b; i++) {cnt = res = 0, n = i;for(int j = 2; j * j <= n; j++)if(n % j == 0) {p[cnt++] = j;while(n % j == 0) n/=j;}if(n > 1) p[cnt++] = n;dfs(0, cnt, 1, 1);ans += res;}printf("Case %d: %I64d\n", __, ans);}int main(void){int _, c, d, k, __;init();while(scanf("%d", &_)!=EOF) {__ = 0;while(_--) {scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);if(k == 0) {printf("Case %d: 0\n", ++__);continue;}a = b/k, b = d/k;work(++__);}}return 0;}

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