思路：
判断一个数列是否是等差数列：
1，最大值减去最小值==（区间个数-1）*k；
2，gcd==k；
3，不能有重复（不会这判断这一条，但是数据水就过了）；
来，上代码：
#include#include #include #include using namespace std; #define maxn 300005 struct TreeNodeType { int l, r, max, min, gcd, mid;};struct TreeNodeType tree[maxn << 2]; int n, m, ai[maxn], ty; inline void in(int &now){ char Cget = getchar(); now = 0; while (Cget > '9' || Cget < '0') Cget = getchar(); while (Cget >= '0'&&Cget <= '9') { now = now * 10 + Cget - '0'; Cget = getchar(); }} int gcd(int a, int b){ int tmp; while(b!=0) tmp=b,b=a%b,a=tmp; return a;} void tree_build(int now, int l, int r){ tree[now].l = l, tree[now].r = r; if (l == r) { tree[now].gcd = ai[l] - ai[l - 1]; tree[now].min = ai[l], tree[now].max = ai[l]; return; } tree[now].mid = l + r >> 1; tree_build(now << 1, l, tree[now].mid); tree_build(now << 1 | 1, tree[now].mid + 1, r); tree[now].gcd = gcd(tree[now << 1].gcd, tree[now << 1 | 1].gcd); tree[now].max = max(tree[now << 1].max, tree[now << 1 | 1].max); tree[now].min = min(tree[now << 1].min, tree[now << 1 | 1].min);} void pre(){ in(n), in(m); for (int i = 1; i <= n; i++) in(ai[i]); tree_build(1, 1, n);} int tree_do(int now, int l, int r){ if (tree[now].l == l&&tree[now].r == r) { if (ty == 1) return tree[now].max; if (ty == 2) return tree[now].min; if (ty == 3) return tree[now].gcd; tree[now].gcd = ai[l] - ai[l - 1]; tree[now].max = ai[l], tree[now].min = ai[l]; return 0; } int res=0; if (l > tree[now].mid) res = tree_do(now << 1 | 1, l, r); else if (r <= tree[now].mid) res = tree_do(now << 1, l, r); else { res = tree_do(now << 1, l, tree[now].mid); if (ty == 1) res = max(res, tree_do(now << 1 | 1, tree[now].mid + 1, r)); if (ty == 2) res = min(res, tree_do(now << 1 | 1, tree[now].mid + 1, r)); if (ty == 3) res = gcd(res, tree_do(now << 1 | 1, tree[now].mid + 1, r)); } if (ty == 4) { tree[now].gcd = gcd(tree[now << 1].gcd, tree[now << 1 | 1].gcd); tree[now].max = max(tree[now << 1].max, tree[now << 1 | 1].max); tree[now].min = min(tree[now << 1].min, tree[now << 1 | 1].min); } return res;} void solve(){ int op, l, r, x, y, last = 0; for (int t = 1; t <= m; t++) { in(op); if (op == 1) { in(x), in(y); x^=last,y^=last,ty=4,ai[x]=y; tree_do(1,x,x); if(x!=n) tree_do(1,x+1,x+1); } else { in(l),in(r),in(x); l^=last,r^=last,x^=last; if(l==r) { last++; printf("Yes\n"); continue; } int ma,mi,gcdd; ty=1,ma=tree_do(1,l,r); ty=2,mi=tree_do(1,l,r); ty=3,gcdd=tree_do(1,l+1,r); if((ma-mi)==x*(r-l)&&abs(gcdd)==x) printf("Yes\n"),last++; else printf("No\n"); } }} int main(){ pre(); solve(); return 0;}