Gym 101201J Shopping(RMQ +二分 )

思路：因为每次除余一个数，那么每次至少减少一半，发现这个后就可以直接暴力。例如对于a， 每次找区间中第一个小于等于a的数进行除余。用RMQ+二分时间复杂度（qlogloglogn）.

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + 10;
int n, q;
LL d[maxn << 2][20];
LL a[maxn];
void RMQ_init(){
for(int i = 0; i < n; ++i) d[i][0] = a[i];
for(int j = 1; (1 << j) <= n; ++j){
for(int i = 0; i + (1 << j) - 1 < n; ++i){
d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}
}
}
LL RMQ(int L, int R){
int k = 0;
while((1 << (k + 1)) <= R - L + 1) k++;
return min(d[L][k], d[R - (1 << k) + 1][k]);
}
int main(){
while(scanf("%d%d", &n, &q) == 2){
for(int i = 0; i < n; ++i)
scanf("%I64d", &a[i]);
RMQ_init();
for(int i = 0; i < q; ++i){
int l, r;
LL v;
scanf("%I64d%d%d", &v, &l, &r);
l--, r--;
int pos = l;
while(pos <= r){
int L = pos, R = r;
int tt = -1;
while(L <= R){
int mid = (L + R) >> 1;
LL t = RMQ(L, mid);
if(t == v){
v = 0;
break;
}
if(t < v){
tt = mid;
R = mid - 1;
}
else L = mid + 1;
}
if(tt == -1) break;
pos = tt + 1;
if(v == 0) break;
v %= a[tt];
}
printf("%I64d\n", v);
}
}
return 0;
}