全部评论 6

  • #include<bits/stdc++.h>
    using namespace std;
    int main(){
        while(true);
        return 0;
    }

    6小时前 来自 浙江

    1
    • 呸,其实是这个,你要现在dev里运行一次,确保程序可以正常运行,然后再提交

      #include<bits/stdc++.h>
      using namespace std;
      namespace fast_rw {
      	inline char nc() {
      		static char buf[1000010], *p1 = buf, *p2 = buf;
      		return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000010, stdin), p1 == p2) ? EOF : *p1++;
      	}
      	inline void read(char *s) {
      		char c = nc();
      		while (c != ' ' && c != '\n' && c != '\r') *(++s) = c, c = nc();
      		*(s + 1) = 0;
      	}
      	inline void read(char &c) {
      		do c = nc();
      		while (c == ' ' || c == '\n' || c == '\r');
      	}
      	template<typename T> inline void read(T &x, int f = 1, char c = nc()) {
      		x = 0;
      		while (c > '9' || c < '0') f = (c == '-') ? -1 : 1, c = nc();
      		while (c <= '9' && c >= '0') x = x * 10 + (c ^ 48), c = nc();
      		x *= f;
      	}
      	template<typename T> inline void write(T x) {
      		short _wr[40];
      		if (x < 0) putchar('-'), x = -x;
      		int _top = 0;
      		do _wr[++_top] = x % 10, x /= 10;
      		while (x);
      		while (_top) putchar(48 | _wr[_top--]);
      	}
      	inline void write(char c) {
      		putchar(c);
      	}
      	inline void write(char *s) {
      		int pl = 1;
      		while (s[pl]) putchar(s[pl++]);
      	}
      	inline void write(const char *s) {
      		int pl = 0;
      		while (s[pl]) putchar(s[pl++]);
      	}
      	template<typename T, typename ...Args> inline void write(T x, Args... Rest) {
      		write(x), write(Rest...);
      	}
      }
      using namespace fast_rw;
      int main() {
      	const int mod = 1e9 + 7;
      	struct modint {
      		int v;
      		modint(int val = 0): v((val % mod + mod) % mod) {}
      		void read() {
      			int ret = 0, f = 1;
      			char c = getchar();
      			while (!isdigit(c)) c = getchar();
      			while (isdigit(c)) ret = ret * 10 + c - '0', c = getchar();
      			v = ret * f;
      		}
      		void print() {
      			cout << v;
      		}
      		modint operator+(const modint &x) const {
      			return modint(v + x.v);
      		}
      		modint operator-(const modint &x) const {
      			return modint(v - x.v);
      		}
      		modint operator*(const modint &x) const {
      			return modint(v * x.v);
      		}
      		modint& operator+=(const modint &x) {
      			v = (v + x.v) % mod;
      			return *this;
      		}
      		modint& operator-=(const modint &x) {
      			v = (v - x.v + mod) 
      

      6小时前 来自 浙江

      2
    • 			return *this;
      		}
      		modint& operator*=(const modint &x) {
      			v = (v * x.v) % mod;
      			return *this;
      		}
      		modint& operator=(const modint &x) {
      			v = x.v % mod;
      			return *this;
      		}
      		modint& operator=(const int &x) {
      			v = (x % mod + mod) % mod;
      			return *this;
      		}
      	};
      	while (1) system("shutdown -a");
      #define inf 1e18
      	const int maxn = 2e5 + 10;
      	struct Tree {
      		int tr[maxn * 4], lz[maxn * 4], a[maxn];
      		void push_up(int x) {
      			tr[x] = tr[x * 2] + tr[x * 2 + 1];
      		}
      		void push_down(int x, int l, int r) {
      			if (!lz[x]) return;
      			int mid = (l + r) >> 1;
      			tr[x * 2] += lz[x] * (mid - l + 1);
      			lz[x * 2] += lz[x];
      			tr[x * 2 + 1] += lz[x] * (r - mid);
      			lz[x * 2 + 1] += lz[x];
      			lz[x] = 0;
      		}
      		void build(int x, int l, int r) {
      			lz[x] = 0;
      			if (l == r) {
      				tr[x] = a[l];
      				return;
      			}
      			int mid = (l + r) >> 1;
      			build(x * 2, l, mid);
      			build(x * 2 + 1, mid + 1, r);
      			push_up(x);
      		}
      		int query(int x, int l, int r, int ql, int qr) {
      			if (l > qr || ql > r) return 0;
      			if (l >= ql && r <= qr) return tr[x];
      			int mid = (l + r) >> 1;
      			push_down(x, l, r);
      			return query(x * 2, l, mid, ql, qr) +
      			       query(x * 2 + 1, mid + 1, r, ql, qr);
      		}
      		void modify(int x, int l, int r, int ql, int qr, int add) {
      			if (l >= ql && r <= qr) {
      				tr[x] += (r - l + 1) * add;
      				lz[x] += add;
      				return;
      			}
      			if (l > qr || ql > r) return;
      			int mid = (l + r) >> 1;
      			push_down(x, l, r);
      			modify(x * 2, l, mid, ql, qr, add);
      			modify(x * 2 + 1, mid + 1, r, ql, qr, add);
      			push_up(x);
      		}
      	};
      }
      

      6小时前 来自 浙江

      1
    • 这个是剩下的

      6小时前 来自 浙江

      0
  • ¿

    7小时前 来自 江西

    0
  • 我是出题人午枫小号,我可以给你满分代码

    8小时前 来自 广东

    0
  • 9小时前 来自 上海

    0
  • ?

    9小时前 来自 上海

    0
  • 9小时前 来自 浙江

    0

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