【AtCoder】ARC100 題解
阿新 • • 發佈:2018-10-14
cut swa std while 處理 方式 累加 endif matrix
C - Linear Approximation
找出\(A_i - i\)的中位數作為\(b\)即可
題解
#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(); }
D - Equal Cut
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總體絕對值相差最小,切的位置不斷右移,直接兩個指針掃一遍即可
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define 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; }
E - Or Plus Max
我們轉化一下問題,處理出每個i or j正好是k的子集的答案,再處理成前綴max即可
這樣的話類似FMT的更新每個子集裏的最大和次大即可
#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(); }
F - Colorful Sequences
我不會數數啊QAQ
先求出所有的序列裏M這一段出現的次數的總和
答案是\((N - M + 1)K^{N - M}\)
然後求M這一段出現在不多彩的序列裏次數的總和
如果M已經是多彩的了,那麽答案是0
如果M不是多彩的且沒有重復的數字
那麽求所有N長的序列裏M長含有不同數字的連續子段有多少個,答案除上\(\frac{K!}{(K - M)!}\)
那麽記錄dp[i][j]作為第i個,然後前j個數都是互不相同的數,j+1開始出現重復
更新的時候從dp[i - 1][h]更新
\(\left\{\begin{matrix} 1 & h \geq j \\ K - h & h = j - 1\\ 0 & h < j - 1 \end{matrix}\right.\)
然後用前綴和處理可以做到\(O(NK)\)
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長含有不同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
剩下轉移方式類似
如果M是多彩的且有重復數字
那麽就記錄F為前綴最多到哪是互不相同的,B為後綴最多到哪是互不相同的
枚舉M所在位置用類似的dp轉移
#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();
}
【AtCoder】ARC100 題解