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Sample Output
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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原文地址:http://www.cnblogs.com/zxhl/p/5866300.html