等差区间(写的很脑残)——中国(北方)大学生程序设计训练赛(第一周)F

等差区间

Description

已知一个长度为 n

的数组 a[1],a[2],…,a
，我们进行 q 次询问，每次询问区间 a[l],a[l+1],…,a[r−1],a[r]，数字从小到大排列后，是否会形成等差数列。等差数列的定义为，数列相邻两项（后一项减去前一项）的差值相等。

Input

本题有多组输入数据。

每组输入数据第一行输入两个正整数n和q

第二行输入n个正整数 a[1],a[2],…,a
。最后输入q行，每行两个数字l,r（1≤l≤r≤n），表示询问区间 a[l],…,a[r]。

1≤n,q≤105,1≤a[i]≤106

Output

对于每组询问输出一行，如果形成等差数列，输出“Yes ”，否则输出“No”（不含引号）。

Sample Input

5 5

3 1 5 2 4

1 3

4 5

1 4

3 4

2 2

Sample Output

Yes

Yes

No

Yes

Yes

Source

lsh

我去，简直瞎了,对了半天的文件，最后发现某数组开小了

//============================================================================
// Name        : tree.cpp
// Author      : Qihan
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <bits/stdc++.h>
using namespace std;
#define pi acos(-1.0)
#define Lowbit(x) (x & (-x))
#define lson l,mid,rt << 1
#define rson mid + 1,r,rt << 1 | 1
#define inf 0x3f3f3f
const int maxn = (100000 + 10);
const int mol = 1000000007;
typedef long long int LLI;
typedef pair<int,int> PII;

struct Node{
int l,r;
int mins,maxs,pre;
}tree[maxn << 2];

int a[maxn],p = 0;
int last[maxn * 10];
int b[maxn];

void pushup(int rt){
tree[rt].maxs = max(tree[rt << 1].maxs,tree[rt << 1 | 1].maxs);
tree[rt].mins = min(tree[rt << 1].mins,tree[rt << 1 | 1].mins);
tree[rt].pre = max(tree[rt << 1].pre,tree[rt << 1 | 1].pre);
}

void Build(int l,int r,int rt){
if(l > r)   return;
tree[rt].l = l;
tree[rt].r = r;
if(l == r){
tree[rt].maxs = a[p];
tree[rt].mins = a[p];
tree[rt].pre = b[p];
p ++;
return ;
}
int mid = (l + r) >> 1;
Build(l ,mid,rt << 1);
Build(mid + 1,r,rt << 1 | 1);
pushup(rt);
}

int QueryMax(int l,int r,int rt){
if(l <= tree[rt].l && r >= tree[rt].r){
return tree[rt].maxs;
}
int mid = (tree[rt].l + tree[rt].r) >> 1,maxs = -1;
if(l <= mid)    maxs = max(QueryMax(l,r,rt << 1),maxs);
if(r > mid)     maxs = max(QueryMax(l,r,rt << 1 | 1),maxs);
return maxs;
}

int QueryBMax(int l,int r,int rt){
if(l <= tree[rt].l && r >= tree[rt].r){
return tree[rt].pre;
}
int mid = (tree[rt].l + tree[rt].r) >> 1,maxs = -1;
if(l <= mid)    maxs = max(QueryBMax(l,r,rt << 1),maxs);
if(r > mid)     maxs = max(QueryBMax(l,r,rt << 1 | 1),maxs);
return maxs;
}

int QueryMin(int l,int r,int rt){
if(l <= tree[rt].l && r >= tree[rt].r){
return tree[rt].mins;
}
int mid = (tree[rt].l + tree[rt].r) >> 1,mins = 10000000;
if(l <= mid)    mins = min(mins,QueryMin(l,r,rt << 1));
if(r > mid)     mins = min(mins,QueryMin(l,r,rt << 1 | 1));
return mins;
}

struct Node2{
int l,r;
int Gcd;
}tree2[maxn << 2];

void Build2(int l,int r,int rt){
if(l > r)   return ;
tree2[rt].l = l;
tree2[rt].r = r;
if(l == r){
tree2[rt].Gcd = a[p ++];
return;
}
int mid = (tree2[rt].l + tree2[rt].r) >> 1;
Build2(l,mid,rt <&l
4000
t; 1);
Build2(mid + 1,r, rt << 1 | 1);
tree2[rt].Gcd = __gcd(tree2[rt << 1].Gcd,tree2[rt << 1 | 1].Gcd);
}

int QueryGcd(int l,int r,int rt){
if(l <= tree2[rt].l && r >= tree2[rt].r){
return tree2[rt].Gcd;
}
int mid = (tree2[rt].l + tree2[rt].r) >> 1,Gcd = -1;
if(l <= mid)    Gcd = QueryGcd(l,r,rt << 1);
if(r > mid){
int RGcd = QueryGcd(l,r,rt << 1 | 1);
if(Gcd == -1)   Gcd = RGcd;
else            Gcd = __gcd(Gcd,RGcd);
}
return Gcd;
}

int main() {
//    freopen("/home/qihan/Documents/in1","r",stdin);
//  freopen("/home/qihan/Documents/out1","w",stdout);
int n,m,ca = 1;
while(~scanf("%d%d",&n,&m)){
//      printf("Case : %d\n",ca ++);
//      printf("----------------------------------------\n");
for(int i = 0;i < n;i ++)       scanf("%d",&a[i]);
memset(last,-1,sizeof(last));
for(int i = 0;i < n;i ++){
b[i] = last[a[i]];
last[a[i]] = i + 1;
}
p = 0;
Build(1,n,1);
for(int i = 0;i < n - 1;i ++)   a[i] = abs(a[i + 1] - a[i]);
p = 0;
Build2(1,n - 1,1);
//      printf("----\n");
for(int i = 1;i <= m;i ++){
//            printf("\t%d ",i);
int l,r;
scanf("%d%d",&l,&r);
if(l == r){
printf("Yes\n");
continue;
}
int maxs = QueryMax(l,r,1);
int mins = QueryMin(l,r,1);
int Bmax = QueryBMax(l,r,1);
int Gcd = QueryGcd(l,r - 1,1);
if(maxs == mins)    printf("Yes\n");
else if(Bmax < l && maxs - mins == (LLI)Gcd * (r - l))  printf("Yes\n");
else            printf("No\n");
}
}
return 0;
}

测试了半天的数据

以及答案

QaQ