找出\(A_i - i\)的中位數做爲\(b\)便可ios
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #define enter putchar('\n') #define space putchar(' ') #define fi first #define se second #define mp make_pair #define MAXN 200005 //#define ivorysi #define pii pair<int,int> using namespace std; typedef long long int64; template<class T> void read(T &res) { res = 0;char c = getchar();T f = 1; 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) {putchar('-');x = -x;} if(x >= 10) out(x / 10); putchar('0' + x % 10); } int64 A[MAXN]; int N; void Solve() { read(N); for(int i = 1 ; i <= N ; ++i) { read(A[i]); A[i] -= i; } sort(A + 1,A + N + 1); int64 t = A[N / 2 + 1],ans = 0; for(int i = 1 ; i <= N ; ++i) { ans += abs(A[i] - t); } out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
枚舉2 和 3中間的位置,兩邊都必須切成絕對值相差最小才能使整體絕對值相差最小,切的位置不斷右移,直接兩個指針掃一遍便可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 enter putchar('\n') #define space putchar(' ') //#define ivorysi #define MAXN 200005 typedef long long int64; using namespace std; template<class T> void read(T &res) { res = 0;char c = getchar();T f = 1; 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[MAXN],sum[MAXN]; int64 get_abs(int l1,int r1,int l2,int r2) { return abs((sum[r1] - sum[l1 - 1]) - (sum[r2] - sum[l2 - 1])); } void Solve() { read(N); for(int i = 1 ; i <= N ; ++i) { read(a[i]);sum[i] = sum[i - 1] + a[i]; } int l = 1,p = 2,r = p + 1; int64 ans = sum[N]; while(p <= N - 2) { while(l + 1 < p && get_abs(1,l,l + 1,p) > get_abs(1,l + 1,l + 2,p)) ++l; r = max(r,p + 1); while(r + 1 < N && get_abs(p + 1,r,r + 1,N) > get_abs(p + 1,r + 1,r + 2,N)) ++r; int64 m[] = {sum[l],sum[p] - sum[l],sum[r] - sum[p],sum[N] - sum[r]}; int64 tmp = 0; for(int i = 0 ; i <= 3 ; ++i) { for(int j = i + 1 ; j <= 3 ; ++j) { tmp = max(tmp,abs(m[j] - m[i])); } } ans = min(ans,tmp); ++p; } out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
咱們轉化一下問題,處理出每一個i or j正好是k的子集的答案,再處理成前綴max便可app
這樣的話相似FMT的更新每一個子集裏的最大和次大便可spa
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <cmath> #include <queue> #include <ctime> #include <map> #include <set> #define fi first #define se second #define pii pair<int,int> //#define ivorysi #define mp make_pair #define pb push_back #define enter putchar('\n') #define space putchar(' ') #define MAXN 100005 using namespace std; typedef long long int64; typedef double db; typedef unsigned int u32; 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 - '0' + c; 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; pii f[(1 << 18) + 5]; int ans[(1 << 18) + 5]; pii Merge(pii a,pii b) { if(a.fi < b.fi) swap(a,b); return mp(a.fi,max(a.se,b.fi)); } void Solve() { read(N); for(int i = 0 ; i < (1 << N) ; ++i) { read(f[i].fi);f[i].se = 0; } for(int i = 1 ; i < (1 << N) ; i <<= 1) { for(int j = 0 ; j < (1 << N) ; ++j) { if(j & i) { f[j] = Merge(f[j],f[j ^ i]); } } } for(int i = 1 ; i < (1 << N) ; ++i) { ans[i] = f[i].fi + f[i].se; ans[i] = max(ans[i - 1],ans[i]); out(ans[i]);enter; } } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
我不會數數啊QAQ指針
先求出全部的序列裏M這一段出現的次數的總和
答案是\((N - M + 1)K^{N - M}\)
而後求M這一段出如今很少彩的序列裏次數的總和code
若是M已是多彩的了,那麼答案是0get
若是M不是多彩的且沒有重複的數字
那麼求全部N長的序列裏M長含有不一樣數字的連續子段有多少個,答案除上\(\frac{K!}{(K - M)!}\)
那麼記錄dp[i][j]做爲第i個,而後前j個數都是互不相同的數,j+1開始出現重複
更新的時候從dp[i - 1][h]更新
\(\left\{\begin{matrix} 1 & h \geq j \\ K - h & h = j - 1\\ 0 & h < j - 1 \end{matrix}\right.\)
而後用前綴和處理能夠作到\(O(NK)\)
用cnt[i][j]表示i長的序列裏,倒數j個數都是互不相同的數,M長含有不一樣數字的連續子段有多少個
在每遇到一個dp[i][j]的j>=M就累加上
剩下轉移方式相似string
若是M是多彩的且有重複數字
那麼就記錄F爲前綴最多到哪是互不相同的,B爲後綴最多到哪是互不相同的
枚舉M所在位置用相似的dp轉移it
#include <iostream> #include <cstdio> #include <algorithm> #include <vector> #include <cstring> //#define ivorysi #define fi first #define se second #define MAXN 25005 #define enter putchar('\n') #define space putchar(' ') typedef long long ll; using namespace std; template <class T> void read(T &res) { res = 0;char c = getchar();T f = 1; while(c < '0' || c > '9') { c = getchar(); if(c == '-') f = -1; } 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;} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } const int MOD = 1000000007; int N,K,M; int A[MAXN],fac[MAXN],invfac[MAXN],inv[MAXN]; int F,B,L,vis[405]; int dp[MAXN][405],cnt[MAXN][405],sum[405],sum_cnt[405],f[MAXN],b[MAXN]; int mul(int a,int b) { return 1LL * a * b % MOD; } int inc(int a,int b) { return a + b >= MOD ? a + b - MOD : a + b; } 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; } void Init() { read(N);read(K);read(M); for(int i = 1 ; i <= M ; ++i) read(A[i]); inv[1] = 1; for(int i = 2 ; i <= K ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i); fac[0] = invfac[0] = 1; for(int i = 1 ; i <= K ; ++i) { fac[i] = mul(fac[i - 1],i); invfac[i] = mul(invfac[i - 1],inv[i]); } F = 0;B = 0; memset(vis,0,sizeof(vis)); for(int i = 1 ; i <= M ; ++i) { if(!vis[A[i]]) { ++F; vis[A[i]] = 1; } else break; } memset(vis,0,sizeof(vis)); for(int i = M ; i >= 1 ; --i) { if(!vis[A[i]]) { ++B; vis[A[i]] = 1; } else break; } memset(vis,0,sizeof(vis)); int l = 0; for(int i = 1 ; i <= M ; ++i) { l = max(l,vis[A[i]]); L = max(L,i - l); vis[A[i]] = i; } } void Process(int st,int *a) { memset(dp,0,sizeof(dp)); dp[0][st] = 1; memset(sum,0,sizeof(sum)); for(int i = st ; i <= K ; ++i) sum[i] = 1; a[0] = 1; for(int i = 1 ; i <= N ; ++i) { for(int j = 1 ; j < K ; ++j) { dp[i][j] = inc(dp[i][j],mul(dp[i - 1][j - 1],(K - j + 1))); dp[i][j] = inc(dp[i][j],inc(sum[K],MOD - sum[j - 1])); } for(int j = 1 ; j <= K ; ++j) { sum[j] = inc(sum[j - 1],dp[i][j]); } a[i] = sum[K - 1]; } } void Solve() { if(L == K) { out(mul(N - M + 1,fpow(K,N - M)));enter; } else if(F == M) { dp[0][0] = 1; int ans = mul(N - M + 1,fpow(K,N - M)),tmp = 0; for(int i = 1 ; i <= N ; ++i) { for(int j = 1 ; j < K ; ++j) { dp[i][j] = inc(dp[i][j],mul(dp[i - 1][j - 1],(K - j + 1))); dp[i][j] = inc(dp[i][j],inc(sum[K],MOD - sum[j - 1])); cnt[i][j] = inc(cnt[i][j],mul(cnt[i - 1][j - 1],(K - j + 1))); cnt[i][j] = inc(cnt[i][j],inc(sum_cnt[K],MOD - sum_cnt[j - 1])); if(j >= M) cnt[i][j] = inc(cnt[i][j],dp[i][j]); } for(int j = 1 ; j <= K ; ++j) { sum[j] = inc(sum[j - 1],dp[i][j]); sum_cnt[j] = inc(sum_cnt[j - 1],cnt[i][j]); } } for(int j = 0 ; j < K ; ++j) { tmp = inc(tmp,cnt[N][j]); } tmp = mul(tmp,fpow(mul(fac[K],invfac[K - M]),MOD - 2)); ans = inc(ans,MOD - tmp); out(ans);enter; } else { Process(F,f);Process(B,b); int ans = mul(N - M + 1,fpow(K,N - M)); for(int i = 1 ; i <= N - M + 1 ; ++i) { int j = i + M - 1; ans = inc(ans,MOD - mul(f[i - 1],b[N - j])); } out(ans);enter; } } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Init(); Solve(); }