c/c++语言开发共享cf896C. Willem, Chtholly and Seniorious(ODT)

题意

题目链接

sol

odt板子题。就是用set维护连续段的大暴力。。

然鹅我抄的板子本题re提交ac？？。。

具体来说，用50 50 658073485 946088556这个数据测试下面的代码，然后在79行停住，看一下bg和i的值会发生神奇的事情。。

问题已解决，确实是那位博主写错了， 只要把split(l)和split(r + 1)反过来就行了

#include<bits/stdc++.h> #define ll long long  #define int long long  #define sit set<node>::iterator  #define fi first #define se second  using namespace std; const int mod = 1e9 + 7, maxn = 1e6 + 10; inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int n, m, mod; template<typename a, typename b> inline int mul(a x, b y) {     return 1ll * x * y % mod; } template<typename a, typename b> inline void add2(a &x, b y) {     x = (x + y % mod); } ll seed, vmax; int a[maxn]; int rnd() {     int ret = seed;     seed = (seed * 7 + 13) % mod;     return ret; } struct node {     int l, r;     mutable ll v;     bool operator < (const node &rhs) const {         return l < rhs.l;     } }; set<node> s; sit split(int p) {     auto pos = s.lower_bound({p});     if(pos != s.end() && pos->l == p) return pos;      pos--;     int l = pos->l, r = pos->r, v = pos->v;     s.erase(pos);     s.insert({l, p - 1, v});      return s.insert({p, r, v}).fi; } void add(int l, int r, int val) {     auto bg = split(l), ed = split(r + 1);     for(auto i = bg; i != ed; i++) i->v += val; } void mem(int l, int r, int val) {     for(int i = l; i <= r; i++) a[i] = val;     auto bg = split(l), ed = split(r + 1);     s.erase(bg, ed);     s.insert({l, r, val}); } int rak(int l, int r, int x) {     vector<pair<ll, int>> v;     auto bg = split(l), ed = split(r + 1);     for(auto i = bg; i != ed; i++)          v.push_back({i->v, i->r - i->l + 1});         sort(v.begin(), v.end());     for(auto it = v.begin(); it != v.end(); it++) {         x -= it -> se;         if(x <= 0) return it -> fi;     }     assert(0); } int fp(int a, int p) {     int base = 1; a %= mod;     while(p) {         if(p & 1) base = 1ll * base * a % mod;         a = 1ll * a * a % mod; p >>= 1;     }     return base; } int po(int l, int r, int x) {     if(l == 6 && r == 7) {         puts("g");     }     auto bg = split(l);     printf("%d %d %d %dn", bg->l, bg->r, bg->v);     auto ed = split(r + 1);     int ans = 0;     for(sit i = bg; i != ed; i++) {         printf("%d %d %d %dn", i->l, i->r, i->v);         ans = (ans + (i->r - i->l + 1) * fp(i->v, x) % mod) % mod;     }     return ans; } signed main() {     n = read(); m = read(); seed = read(); vmax = read();     for(int i = 1; i <= n; i++) a[i] = (rnd() % vmax) + 1, s.insert({i, i, a[i]});     s.insert({n + 1, n + 1, 0});     for(int i = 1; i <= m; i++) {         int op = (rnd() % 4) + 1; int l = (rnd() % n) + 1; int r = (rnd() % n) + 1, x = -1;         if(l > r) swap(l, r);                  if(op == 3) x = (rnd() % (r - l + 1)) + 1;         else x = (rnd() % vmax) + 1;         if(op == 4) mod = (rnd() % vmax) + 1;          if(op == 1) add(l, r, x);         else if(op == 2) mem(l, r, x);         else if(op == 3) cout << rak(l, r, x) << 'n';         else cout << po(l, r, x) << 'n';     }     return 0; } /* 50 50 658073485 946088556 */