GCD Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1233 Accepted Submission(s): 382 Problem Description Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to calculate gcd(al,,al+1,...,ar) and count the number of pairs(l′,r′)(1≤l<r≤N)such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar). Input The first line of input contains a number T, which stands for the number of test cases you need to solve. The first line of each case contains a number N, denoting the number of integers. The second line contains N integers, a1,...,an(0<ai≤1000,000,000). The third line contains a number Q, denoting the number of queries. For the next Q lines, i-th line contains two number , stand for the li,ri, stand for the i-th queries. Output For each case, you need to output “Case #:t” at the beginning.(with quotes, t means the number of the test case, begin from 1). For each query, you need to output the two numbers in a line. The first number stands for gcd(al,al+1,...,ar) and the second number stands for the number of pairs(l′,r′) such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar). Sample Input 1 5 1 2 4 6 7 4 1 5 2 4 3 4 4 4 Sample Output Case #1: 1 8 2 4 2 4 6 1
/* 难点在于计数,关键要发现gcd的递减性 对于1到n的(l,r) 对于固定的l gcd(l,r)>=gcd(l,r+1) 对于固定的r gcd(l,r)<=gcd(l+1,r) 因此可以不用逐一地进行计数 方法为: 枚举每一个左区间,对满足gcd(l,ri)的右区间进行二分查找 跳跃着进行计数 */ #include <algorithm> #include <cstring> #include <cstdio> #include <iostream> #include <map> #include <vector> #define scan1(x) scanf("%d",&x) #define scan2(x,y) scanf("%d%d",&x,&y) #define scan3(x,y,z) scanf("%d%d%d",&x,&y,&z) using namespace std; typedef long long LL; const int inf=0x3f3f3f3f; const int Max=1e6+10; int dp[Max][30]; map<int,LL> vis; int A[Max]; int gcd(int x,int y) { if(x<y) swap(x,y); return (y==0?x:gcd(y,x%y)); } void RMQ_init(int n) { for(int i=1; i<=n; i++) dp[i][0]=A[i]; for(int j=1; (1<<j)<=n; j++) { for(int i=1; i+(1<<j)-1<=n; i++) { dp[i][j]=gcd(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); } } } int RMQ(int l,int r) { int k=0; while((1<<(k+1))<=r-l+1) k++; return gcd(dp[l][k],dp[r-(1<<k)+1][k]); } void Init() { vis.clear(); } int L[Max],R[Max]; int main() { int T,ca=1; for(scan1(T); T; T--) { int n,m,num; scan1(n); for(int i=1; i<=n; i++) scan1(A[i]); RMQ_init(n); Init(); scan1(m); for(int i=1; i<=m; i++) { scan2(L[i],R[i]); num=RMQ(L[i],R[i]); vis.insert(make_pair(num,0)); } int l,r,mid,d1,d2,ans,nex; for(int i=1; i<=n; i++) { nex=i; while(nex<=n) { d1=RMQ(i,nex); l=nex;r=n; while(l<=r) { mid=(l+r)>>1; d2=RMQ(i,mid); if(d2>=d1) l=mid+1,ans=mid; else r=mid-1; } if(vis.find(d1)!=vis.end()) vis[d1]+=(ans-nex)+1; nex=r+1; } } printf("Case #%d:\n",ca++); for(int i=1; i<=m; i++) { ans=RMQ(L[i],R[i]); printf("%d %lld\n",ans,vis[ans]); } } return 0; }
时间: 2024-10-07 05:24:51