就是线段树板子题，还是单点修改区间查询。
用一个指针cnt记录当前序列里有几个数，然后操作1就是把++cnt的位置的数改为(n + t) % d；操作2就是查询cnt - L + 1到cnt的区间最大值。
我用的是先把线段树的节点开好的方法，所以这题按区间长度等于m开就行。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter printf(" ") 13 #define space printf(" ") 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const int eps = 1e-8; 19 const int maxn = 2e5 + 5; 20 inline ll read() 21 { 22 ll ans = 0; 23 char ch = getchar(), last = ' '; 24 while(!isdigit(ch)) {last = ch; ch = getchar();} 25 while(isdigit(ch)) 26 { 27 ans = ans * 10 + ch - '0'; ch = getchar(); 28 } 29 if(last == '-') ans = -ans; 30 return ans; 31 } 32 inline void write(ll x) 33 { 34 if(x < 0) x = -x, putchar('-'); 35 if(x >= 10) write(x / 10); 36 putchar(x % 10 + '0'); 37 } 38 39 int m, cnt = 0; 40 ll mod; 41 ll t = 0; 42 43 int l[maxn << 2], r[maxn << 2]; 44 ll Max[maxn << 2]; 45 void build(int L, int R, int now) 46 { 47 l[now] = L; r[now] = R; 48 if(L == R) return; 49 int mid = (L + R) >> 1; 50 build(L, mid, now << 1); 51 build(mid + 1, R, now << 1 | 1); 52 } 53 void update(int id, ll d, int now) 54 { 55 if(l[now] == r[now]) {Max[now] = d; return;} 56 int mid = (l[now] + r[now]) >> 1; 57 if(id <= mid) update(id, d, now << 1); 58 else update(id, d, now << 1 | 1); 59 Max[now] = max(Max[now << 1], Max[now << 1 | 1]); 60 } 61 ll query(int L, int R, int now) 62 { 63 if(l[now] == L && r[now] == R) return Max[now]; 64 int mid = (l[now] + r[now]) >> 1; 65 if(R <= mid) return query(L, R, now << 1); 66 else if(L > mid) return query(L, R, now << 1 | 1); 67 else return max(query(L, mid, now << 1), query(mid + 1, R, now << 1 | 1)); 68 } 69 70 int main() 71 { 72 m = read(); mod = read(); 73 build(1, m, 1); 74 for(int i = 1; i <= m; ++i) 75 { 76 char c; cin >> c; 77 if(c == 'A') 78 { 79 ll n = read(); 80 update(++cnt, (n + t) % mod, 1); 81 } 82 else 83 { 84 int L = read(); 85 t = query(cnt - L + 1, cnt, 1); 86 write(t); enter; 87 } 88 } 89 return 0; 90 }