中国（北方）大学生程序设计训练赛（第一周）(Problem F: 等差区间-线段树＋等差数列平方和公式)

对一长度为105的静态序列进行q次询问，每次询问某段区间元素是否排序后能成为等差数列

正解：
考虑2种情况
公差为0,
公差不为0，此时求区间gcd，区间内无相同元素 且 相邻两项差构成的数列的GCD *(R-L) = (区间最大值-区间最小值）

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <map>#include <functional>#include <cstdlib>#include <queue>#include <stack>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase) printf("Case %d:\n",kcase);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} const int maxn =1000000;ull sum[maxn],a[maxn],minv[maxn],maxv[maxn],a2[maxn],sum2[maxn];void pushup(int o) {    sum[o]=sum[Lson]+sum[Rson];    sum2[o]=sum2[Lson]+sum2[Rson];    maxv[o]=max(maxv[Lson],maxv[Rson]);    minv[o]=min(minv[Lson],minv[Rson]);}void build(int l,int r,int o) {    if (l==r) {        minv[o]=maxv[o]=sum[o]=a[l];        sum2[o]=(ull)a[l]*a[l];        return ;    }    int m=(l+r)>>1;    build(l,m,Lson),build(m+1,r,Rson);    pushup(o);}ull mi1,ma1,s1,s2;void query(int l,int r,int o,int L,int R) {    if(L<=l && r<=R ) {        mi1=min(mi1,minv[o]);        ma1=max(ma1,maxv[o]);        s1+=sum[o];        s2+=sum2[o];        return;    }    int m=(l+r)>>1;    if(L<=m) query(l,m,Lson,L,R);    if(m<R) query(m+1,r,Rson,L,R); }int main(){//  freopen("F.in","r",stdin);//  freopen(".out","w",stdout);    int n,q;    while(scanf("%d%d",&n,&q)==2) {         For(i,n) a[i]=read();        build(1,n,1);        For(i,q) {            int l=read(),r=read();            mi1=INF,ma1=0,s1=s2=0;            query(1,n,1,l,r);            ull len=r-l+1;            bool flag=0;            if (len==1) flag=1;            else {                if ((ma1-mi1)%(len-1)!=0) flag=0;                else {                    ull d=ma1-mi1;                    d/=len-1;                    ull ss1=(mi1+ma1)*len;                    if (ss1!=s1*2) {flag=0;}                    else {                        ull ss2=len*mi1*mi1;                        ss2+=len*(len-1)/2*(2*len-1)/3*d*d;                        ss2+=len*(len-1)*d*mi1;                        if (ss2==s2) flag=1;else flag=0;                     }                }            }            puts(flag? "Yes":"No");        }    }       return 0;}