HDU 5875 Function

Function

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1095 Accepted Submission(s): 420


[align=left]Problem Description[/align]
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).

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each query(l,r), output F(l,r) on one line.

[align=left]Sample Input[/align]

1

3
2 3 3

1

1 3

[align=left]Sample Output[/align]

2

[align=left]Source[/align]
2016 ACM/ICPC Asia Regional Dalian Online

解析：题意为对于每个询问区间[l, r]，求a[l]%a[l+1]%a[l+2]%...%a[r]。因为当后面的数比前面的数大时，取模得到的结果没有变化。当后面的数比前面的数小或相等时，取模得到的结果才有变化，并且结果越来越小。因此可以进行预处理得到每个数第一个不大于这个数的位置，每次取模时直接跳转到此位置进行。

#include <cstdio>
#include <cstring>

const int MAXN = 1e5+5;
int a[MAXN];
int next[MAXN]; //第一个小于等于这个数的位置
int n;

void solve()
{
memset(next, -1, sizeof(next)); //初始化为-1
for(int i = 1; i <= n; ++i){
for(int j = i+1; j <= n; ++j){
if(a[j] <= a[i]){
next[i] = j;
break;
}
}
}
int q, l, r;
scanf("%d", &q);
while(q--){
scanf("%d%d", &l, &r);
int res = a[l];
for(int i = next[l]; i <= r; i = next[i]){
if(i == -1) //表明后面的数都比它大
break;
if(res == 0)    //0取模后结果还是0
break;
res %= a[i];
}
printf("%d\n", res);
}
}

int main()
{
int t;
scanf("%d", &t);
while(t--){
scanf("%d", &n);
for(int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
solve();
}
return 0;
}