HDU 5875 Function st + 二分
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Function
Problem Description
The shorter, the simpler. With this problem, you should be convinced of this truth.
You are given an array A of N postive integers, and M queries in the form (l,r). A function F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.
You job is to calculate F(l,r), for each query (l,r).
You are given an array A of N postive integers, and M queries in the form (l,r). A function F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.
You job is to calculate F(l,r), for each query (l,r).
Input
There are multiple test cases.
The first line of input contains a integer T, indicating number of test cases, and T test cases follow.
For each test case, the first line contains an integer N(1≤N≤100000).
The second line contains N space-separated positive integers: A1,…,AN (0≤Ai≤109).
The third line contains an integer M denoting the number of queries.
The following M lines each contain two integers l,r (1≤l≤r≤N), representing a query.
The first line of input contains a integer T, indicating number of test cases, and T test cases follow.
For each test case, the first line contains an integer N(1≤N≤100000).
The second line contains N space-separated positive integers: A1,…,AN (0≤Ai≤109).
The third line contains an integer M denoting the number of queries.
The following M lines each contain two integers l,r (1≤l≤r≤N), representing a query.
Output
For each query(l,r), output F(l,r) on one line.
Sample Input
1
3
2 3 3
1
1 3
Sample Output
2
题意:
给你一个n,n个数
m个询问,每次询问你 l,r,, a[l] % a[l+1] % a[l+2] %……a[r] 结果是多少
题解;
每次有效的取模会使结果减半,因此只有log次有效取模,每次往右找一个不大于结果的最靠左的数,ST表+二分
注意RMQ查询的时候少用 log函数,这是造成我开始超时的原因
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<vector> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; const long long INF = 1e18; const double Pi = acos(-1.0); const int N = 1e5+10, M = 1e2+11, mod = 1e9+7, inf = 2e9; int dp[N][50],a[N],n,q; void st() { for(int j = 1; (1<<j) <= n; ++j) { for(int i = 1; (i + (1<<j) - 1) <= n; ++i) { dp[i][j] = min(dp[i][j-1],dp[i + (1<<(j-1))][j-1]); } } } int query(int l,int r) { int len = r - l + 1; int k = 0; while ((1 << (k + 1)) <= len) k++; return min(dp[l][k],dp[r - (1<<k) + 1][k]); } int _binary_search(int l,int r,int res) { int s = r+1; while(l <= r) { int md = (l + r) >> 1; if(query(l,md) <= res) r = md - 1,s = md; else l = md + 1; } return s; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d",&n); for(int i = 1; i <= n; ++i) scanf("%d",&a[i]),dp[i][0]=a[i]; st(); scanf("%d",&q); for(int i = 1; i <= q; ++i) { int x,y,L,R; scanf("%d%d",&x,&y); int res = a[x]; L = x+1, R = y; while(L <= R && res) { L = _binary_search(L,R,res); if(L<=R) { res%=a[L];L++; } } printf("%d\n",res); } } return 0; }
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