标签:-- spl show math ++ 答案 close ring 标记
OJ 1177 1178 1179
一.电路图A
第一问,容易看出$右拐次数=左拐次数+4$,$左拐+右拐=n$,所以$右拐=n/2-2$,相当于$C_{n}^{n/2-2}$
第二问,总个数除去最左端,最右端,最上方,最下方的电阻,整个电路被分成$4$段,每一段都有偶数个电阻,答案就是将$n$分成$4$个偶数的情况,相当于将$n/2$分成任意数的情况,用隔板法,答案就是$C_{n/2-1}^{3}$,排除旋转的影响,答案还要除以$4$。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <deque> #include <string> using namespace std; inline long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) f *= (ch == ‘-‘) ? -1 : 1, ch = getchar(); do ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar(); while(isdigit(ch)); return ans * f; } const int MOD = 1e9+7, MAXN = 1e7 + 1; int fac[MAXN], inv[MAXN]; int pow(int a,int p){ int ans = 1; while(p){ if(p & 1) ans = (long long)ans * a % MOD; a = (long long)a * a % MOD, p >>= 1; } return ans; } inline int C(int n,int m){ return (long long)fac[n] * inv[m] % MOD * inv[n-m] % MOD; } int main(){ int n = read(), ans1, ans2; fac[0] = 1; for(int i=1; i<=n; i++) fac[i] = (long long)fac[i-1] * i % MOD; inv[n] = pow(fac[n], MOD-2); for(int i=n-1; i>=0; i--) inv[i] = (long long)inv[i+1] * (i+1) % MOD; ans1 = C(n, n/2-2); ans2 = (long long)C(n/2+1, 3) * n % MOD * pow(4, MOD-2) % MOD; printf("%d\n%d", ans1, ans2); return 0; }
二.电路图B
区间修改,区间查询,一看就是线段树,关键在于合并标记如何合并。
可将一个标记记为$\frac{ax+b}{cx+d}$,用四个数表示为$\{a,b,c,d\}$,串联一个电阻就是$\{0,R,0,1\}$,并联一个电阻就是$\{R,0,1,R\}$
最关键是mix操作,将$\{a,b,c,d\}$和$\{i,j,k,p\}$合成为$\{ai+cj,bi+dj,ak+cp,bk+dp\}$,最终就转化为线段树加标记的问题了。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘) f *= (ch == ‘-‘) ? -1 : 1, ch = getchar(); do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar(); while(ch >= ‘0‘ && ch <= ‘9‘); return ans * f; } const int MAXN = 250001, MAXM = 250001; struct sign{double a, b, c, d;}; struct Segment{ int l, r; double maxNum, minNum; sign add; bool isSigned; }seg[MAXN*4]; int a[MAXN]; void change(int x,sign fa){ sign t = seg[x].add; seg[x].isSigned = true; double a, b, c, d; a = t.a * fa.a + t.c * fa.b; b = t.b * fa.a + t.d * fa.b; c = t.a * fa.c + t.c * fa.d; d = t.b * fa.c + t.d * fa.d; seg[x].add = (sign){1, b/a, c/a, d/a}; seg[x].maxNum = (seg[x].maxNum * fa.a + fa.b) / (seg[x].maxNum * fa.c + fa.d); seg[x].minNum = (seg[x].minNum * fa.a + fa.b) / (seg[x].minNum * fa.c + fa.d); } void spread(int x){ if(seg[x].isSigned){ change(x*2, seg[x].add); change(x*2+1, seg[x].add); seg[x].isSigned = false; seg[x].add = (sign){1, 0, 0, 1}; } } void build(int l,int r,int x=1){ seg[x].l = l, seg[x].r = r; seg[x].add = (sign){1, 0, 0, 1}; if(l == r){ seg[x].maxNum = seg[x].minNum = a[l]; return; } int mid = (l + r) >> 1; build(l, mid, x*2); build(mid+1, r, x*2+1); seg[x].maxNum = max(seg[x*2].maxNum, seg[x*2+1].maxNum); seg[x].minNum = min(seg[x*2].minNum, seg[x*2+1].minNum); } void point(int l,int r,sign vis,int x=1){ if(l <= seg[x].l && seg[x].r <= r){ change(x, vis); return; } spread(x); if(seg[x*2].r >= l) point(l, r, vis, x*2); if(seg[x*2+1].l <= r) point(l, r, vis, x*2+1); seg[x].maxNum = max(seg[x*2].maxNum, seg[x*2+1].maxNum); seg[x].minNum = min(seg[x*2].minNum, seg[x*2+1].minNum); } double ask(int l,int r,int type,int x=1){ if(l <= seg[x].l && seg[x].r <= r) return (type == 1) ? seg[x].maxNum : seg[x].minNum; spread(x); double ans = (type == 1) ? 0 : 1e9; if(seg[x*2].r >= l) (type == 1) ? ans = max(ans, ask(l, r, type, x*2)) : ans = min(ans, ask(l, r, type, x*2)); if(seg[x*2+1].l <= r) (type == 1) ? ans = max(ans, ask(l, r, type, x*2+1)) : ans = min(ans, ask(l, r, type, x*2+1)); return ans; } int main(){ int n = read(); for(int i=1; i<=n; i++) a[i] = read(); build(1, n); int m = read(); for(int i=1; i<=m; i++){ int op = read(), l = read(), r = read(); if(op == 1) printf("%.10lf\n", ask(l, r, 1)); else if(op == 2) printf("%.10lf\n", ask(l, r, 2)); else{ int R = read(); if(op == 3) point(l, r, (sign){1, R, 0, 1}); else point(l, r, (sign){R, 0, 1, R}); } } return 0; }
三.电路图C(待填坑)
标签:-- spl show math ++ 答案 close ring 标记
原文地址:https://www.cnblogs.com/PHDHD/p/12276110.html