hdu1796 How many integers can you find(容斥)
阿新 • • 發佈:2018-12-31
題意:給你一個n(n<=2^31),m(m<=10),再給你m個數a[i](0<=a[i]<=20),問你[1,n-1]範圍裡有多少個數能被這m個數中的一個或多個整除。
題解:容斥裸體,注意a[i]=0的情況。
sum = 被1個數整除的個數-被2個數整除的個數+被3個數整除的個數-.... (注意被多個數整除的時候,我們利用的是lcm)
程式碼:
#include<cstdio> #include<cstring> #include<iostream> #include<sstream> #include<algorithm> #include<vector> #include<bitset> #include<set> #include<queue> #include<stack> #include<map> #include<cstdlib> #include<cmath> #define PI 2*asin(1.0) #define LL long long #define pb push_back #define pa pair<int,int> #define clr(a,b) memset(a,b,sizeof(a)) #define lson lr<<1,l,mid #define rson lr<<1|1,mid+1,r #define bug(x) printf("%d++++++++++++++++++++%d\n",x,x) #define key_value ch[ch[root][1]][0]C:\Program Files\Git\bin const LL MOD = 1E9+7; const LL N = 2e5+15; const int maxn = 5e5+15; const int letter = 130; const int INF = 1e17; const double pi=acos(-1.0); const double eps=1e-10; using namespace std; 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*10+ch-'0';ch=getchar();} return x*f; } int n,m,a[15]; LL gcd(LL a,LL b){ if(b==0) return a; return gcd(b,a%b); } LL lcm(LL a,LL b){ return a*b/gcd(a,b); } int main(){ while(scanf("%d%d",&n,&m)!=EOF){ for(int i=0;i<m;i++) { scanf("%d",a+i); if(!a[i])i--,m--; } n--; LL sum=0,ans; int num; for(int i=1;i<(1<<m);i++){ ans=1; num=0; for(int j=0;j<m;j++){ if((1<<j)&i){ num++; ans=lcm(ans,1ll*a[j]); } } if(num%2) sum+=1ll*n/ans; else sum-=1ll*n/ans; } printf("%I64d\n",sum); } return 0; }