#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
#include<ctime>
#include "cstdio"
#include "string"
#include "string.h"
#include "map"
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int st[N][30], n, ans1, a[N];
map<int, ll>mp;
int gcd(int a, int b)
{
return (b>0) ? gcd(b, a%b) : a;
}
void init(int *a,int n)
{
for (int i = 1;i <= n;i++)
st[i][0] = a[i];
for (int j = 1; j<18; j++)
for (int i = 1; i + (1 << j) -1 <= n; i++)
st[i][j] = gcd(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
}
int rmq(int u, int v)
{
int k = 0;
while (u + (1 << k) <= v + 1) k++;
k--;
return gcd(st[u][k], st[v - (1 << k) + 1][k]);
}
ll ans;
int solve(int start,int l,int r,int num)
{
int pos = r + 1;
while (l < r)
{
int mid = (l + r) / 2;
if (rmq(start, mid) < num)
{
r = mid;
pos = r;
}
else
{
l = mid + 1;
pos = l;
}
}
return pos;
}
int main()
{
int T, m, cas = 1;
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
mp.clear();
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
init(a,n);
for (int i = 1;i <= n;i++)
{
for (int j = i;j <= n;)
{
int jump = solve(i, j, n, rmq(i, j));
mp[rmq(i, j)] += jump - j;
j = jump;
}
}
printf("Case #%d:\n", cas++);
scanf("%d", &m);
int a, b;
while (m--)
{
scanf("%d%d", &a, &b);
ans1 = rmq(a, b);
ans = mp[ans1];
printf("%d %lld\n", ans1, ans);
}
}
return 0;
}
2016 Multi-University Training Contest 1 1004 GCD
題目鏈接:點擊打開鏈接
題目大意:給你一串序列,按要求輸出l到r的gcd值,並且找出整條序列中所有gcd值等於l到r的gcd值得子序列數量。
解題思路:一開始沒想明白gcd是否滿足二分性,其實算是滿足的,當開始位置定了之後,隨着長度增加gcd值不上升。同時這題本身需要預處理,但是其實只有logn中因數,那麼枚舉開始點,也就只有nlogn的複雜度,用map保存每種結果的值,st表查詢區間gcd(線段樹會超時)
代碼:
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