這題就是從後面找一個最靠後並且當前能夠放的,能夠放的條件是它的前面正好放了它的數值-1個數node
若是不符合條件就退出c++
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 200005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int a[MAXN],N,cnt[MAXN]; bool vis[MAXN]; void Solve() { read(N); for(int i = 1 ; i <= N ; ++i) read(a[i]); for(int i = 1 ; i <= N ; ++i) { int cnt = 0; for(int j = 1 ; j < i ; ++j) { if(a[j] <= a[i]) ++cnt; } if(cnt + 1 < a[i]) {puts("-1");return;} } for(int i = 1 ; i <= N ; ++i) { cnt[0] = 0; for(int j = 1 ; j <= N ; ++j) cnt[j] = cnt[j - 1] + vis[j]; for(int j = N ; j >= 1 ; --j) { if(!vis[j] && cnt[j - 1] + 1 == a[j]) { vis[j] = 1; out(a[j]);enter; break; } } } } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
若是是偶數個,分紅和相等的\(\frac{N}{2}\)份ui
若是是奇數個,最後一個點單獨爲一組,前\(N - 1\)個兩個一組分紅和相等的\(\frac{N - 1}{2}\)份spa
而後把一組做爲一個點,建一個徹底圖,兩組(一組1,2,一組3,4)之間連邊就是code
1-2,1-4,2-3,2-4blog
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 200005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int N; vector<pii > v; void Solve() { read(N); if(N & 1) { for(int i = 1 ; i < N ; ++i) { v.pb(mp(i,N)); } --N; } for(int i = 1 ; i <= N ; ++i) { for(int j = i + 1 ; j <= N ; ++j) { if(i + j != N + 1) v.pb(mp(i,j)); } } out(v.size());enter; for(auto t : v) { out(t.fi);space;out(t.se);enter; } } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
一定存在一個歐拉回路three
若是一個點有六個或以上點度一定存在ci
若是有三個點以上有四個點度一定存在get
若是隻有兩個點有四個點度it
那麼若是這四條路都在這兩個點之間,那麼就無解
不然有解
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 100005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } struct node { int to,next; }E[MAXN * 2]; int head[MAXN],sumE; int N,M; int cnt[MAXN]; bool vis[MAXN]; void add(int u,int v) { E[++sumE].to = v; E[sumE].next = head[u]; head[u] = sumE; } void dfs(int u) { vis[u] = 1; for(int i = head[u] ; i ; i = E[i].next) { int v = E[i].to; if(!vis[v]) dfs(v); } } void Solve() { read(N);read(M); int a,b; for(int i = 1 ; i <= M ; ++i) { read(a);read(b); cnt[a]++;cnt[b]++; add(a,b);add(b,a); } for(int i = 1 ; i <= N ; ++i) { if(cnt[i] & 1) {puts("No");return;} } int t = 0; int p = 0,q = 0; for(int i = 1 ; i <= N ; ++i) { if(cnt[i] >= 6) {puts("Yes");return;} if(cnt[i] >= 4) { ++t; if(!p) p = i; else if(!q) q = i; } } if(t > 2) {puts("Yes");return;} if(t == 2) { vis[p] = 1; dfs(q); for(int i = 1 ; i <= N ; ++i) { if(!vis[i]) {puts("Yes");return;} } } puts("No"); } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
把操做改爲數軸上,我能夠把一個點移動到左邊或右邊任意一個位置上,能夠沒必要要是整數點
顯然對於每一個點操做只進行一次
而後拆成\((-\infty,1),(1,2),(2,3),(3,4)....(N - 1,N),(N,+\infty)\)和整數點\(1,2,3,4,5,6..N\)
而後根據位置dp便可
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 5005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int N; int64 A,B; int64 dp[MAXN][2 * MAXN],s[MAXN][2 * MAXN]; int p[MAXN],pos[MAXN]; void Solve() { read(N);read(A);read(B); for(int i = 1 ; i <= N ; ++i) {read(p[i]);pos[p[i]] = i;} for(int i = 1 ; i <= N ; ++i) { s[i][0] = 1e18; for(int j = 1 ; j <= 2 * N + 1 ; ++j) { if(j & 1) { dp[i][j] = s[i - 1][j] + (j < pos[i] * 2 ? B : A); } else { dp[i][j] = s[i - 1][j - 1] + (j != pos[i] * 2 ? (j < pos[i] * 2 ? B : A): 0); } s[i][j] = min(s[i][j - 1],dp[i][j]); } } out(s[N][2 * N + 1]);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
用藍色表示相加小於M的,紅色表示大於M的
而後就會變成前面是藍的,後面是紅的
簡單分析發現藍的越小越好,紅的越大越好,因此求出能向左最長的紅色序列
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 200005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int N,M; int a[MAXN]; void Solve() { read(N);read(M); for(int i = 1 ; i <= 2 * N ; ++i) read(a[i]); sort(a + 1,a + 2 * N + 1); int r = -1; for(int i = 2 * N ; i >= 1 ; --i) { int t = lower_bound(a + 1,a + 2 * N + 1,M - a[i]) - a; if((t & 1) == (i & 1)) ++t; if(r == -1) r = i + t; else r = max(r,i + t); } if(r == -1) r = 4 * N + 1; int ans = 0; for(int i = 2 * N ; i >= 1 ; --i) { if(r - i < i) ans = max(ans,(a[i] + a[r - i]) % M); else break; } r = 1 + r - 2 * N - 1; for(int i = 1 ; i <= 2 * N ; ++i) { if(i < r - i) ans = max(ans,a[i] + a[r - i]); else break; } out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
切一刀,畫一條紅線,而後正120畫一條藍線,反120畫一條綠線
咱們要求的就是任意1/3中最短兩個顏色不一樣之間的線的角度大小
而後咱們取第一刀紅線和藍線之間的 1/3區間,而後裏面被分紅了\(N\)份,咱們遞推一個\(f(i)\),表示有\(i\)份左右兩邊區間顏色不一樣的機率
而後\(g(i)\)表示\(i\)份中最短的一段的指望長度,\(\frac{1}{3}\)長度裏\(N\)份中選\(i\)份指望長度是\(\frac{i}{3N}\),能夠認爲分紅\(N\)份中選\(i\)份每一個點被選中的機率是\(\frac{i}{N} = \frac{\binom{i - 1}{N - 1}}{\binom{i}{N}}\)
那\(i\)段中的指望最小長度呢
是\(E(min) = \int_{t = 0}^{1/i}P(min \geq t) dt = \int_{t = 0}^{1/i}(1 - it)^{i - 1}dt = \int_{t = 0}^{1}\frac{t^{i - 1}}{i} dt = \frac{1}{i^2}\)
因此\(g(i) = \frac{1}{3iN}\)
答案就是全部的\(f(i)g(i)\)的和
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 1000005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } const int MOD = 1000000007; int N; int f[MAXN][3],g[MAXN],fac[MAXN],invfac[MAXN],inv[MAXN]; int inc(int a,int b) { return a + b >= MOD ? a + b - MOD : a + b; } int mul(int a,int b) { return 1LL * a * b % MOD; } int fpow(int x,int c) { int res = 1,t = x; while(c) { if(c & 1) res = mul(res,t); t = mul(t,t); c >>= 1; } return res; } int C(int n,int m) { if(n < m) return 0; return mul(fac[n],mul(invfac[m],invfac[n - m])); } void Solve() { read(N); fac[0] = 1; for(int i = 1 ; i <= N; ++i) fac[i] = mul(fac[i - 1],i); invfac[N] = fpow(fac[N],MOD - 2); for(int i = N - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1); inv[1] = 1; for(int i = 2 ; i <= 1000000 ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i); f[0][0] = 1; int iv3= fpow(inv[3],N - 1); int ans = 0; for(int i = 1 ; i <= N ; ++i) { f[i][1] = inc(f[i - 1][0],f[i - 1][2]); f[i][0] = inc(f[i - 1][1],f[i - 1][2]); f[i][2] = inc(f[i - 1][0],f[i - 1][1]); g[i] = mul(mul(inv[3],inv[N]),inv[i]); int p = mul(mul(f[i][1],C(N,i)),iv3); ans = inc(ans,mul(g[i],p)); } out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }