求巅峰赛答案
2025-09-13 10:51:48
发布于：浙江
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我要T6的20分代码！
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#include<bits/stdc++.h> using namespace std; int main(){ while(true); return 0; }2025-09-13 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)2025-09-13 来自 浙江
2return *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); } }; }2025-09-13 来自 浙江
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