【AtCoder】AGC031
A - Colorful Subsequence
答案是node
\(\prod_{c = 'a'}^{'z'} (cnt[c] + 1)\)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 MAXN 200005
#define eps 1e-12
//#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;
char s[MAXN];
int cnt[35];
void Solve() {
read(N);
scanf("%s",s + 1);
for(int i = 1 ; i <= N ; ++i) {
cnt[s[i] - 'a']++;
}
int res = 1;
for(int i = 0 ; i < 26 ; ++i) {
res = 1LL * res * (1 + cnt[i]) % MOD;
}
res = (res + MOD - 1) % MOD;
out(res);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
B - Reversi
記\(dp[i][0/1]\)表示沒有進行操做/進行了操做ui
\(dp[i][0] = dp[i - 1][0] + dp[i - 1][1]\)spa
若是\(C[i - 1] == C[i]\),那麼不能進行操做code
不然能夠從前面全部\(C[k - 1] != C[k]\)且\(C[i] == C[k]\)的地方轉移過來；遞歸
#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 MAXN 200005
#define eps 1e-12
//#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,C[MAXN],sum[MAXN];
int dp[MAXN][2];
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;
}
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(C[i]);
}
dp[0][0] = 1;
for(int i = 1 ; i <= N ; ++i) {
dp[i][0] = inc(dp[i - 1][1],dp[i - 1][0]);
if(C[i] != C[i - 1]) {
dp[i][1] = sum[C[i]];
sum[C[i]] = inc(sum[C[i]],dp[i][0]);
}
}
out(inc(dp[N][0],dp[N][1]));enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
C - Differ by 1 Bit
這道題用數學概括法ip
A和B中1的奇偶性必須不一樣get
顯然N = 1的的時候確定成立數學
若是\(N = K +1\)，且\(N = K\)時成立it
這個時候找\(A\)和\(B\)中不一樣的一位，刪掉，此時\(A\)和\(B\)中1的奇偶性相同
而後如今是兩個\(K\)位數，咱們設一個\(K\)位數\(C\)，和\(A，B\)的1的奇偶性不一樣
而後\(A\)到\(C\)能夠到達，\(C\)到\(B\)也能夠達到，遞歸處理，而後將前\(2^K\)刪掉的位置添加上，且與\(A\)相同，後\(2^{K}\)刪掉的位置添加上和\(B\)相同
#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 MAXN 200005
#define eps 1e-12
//#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,A,B;
int Delete(int x,int p) {
int d = x & (1 << p) - 1;
int u = x & (1 << 20) - (1 << p + 1);
return (u >> 1) + d;
}
int Insert(int x,int p,int v) {
int d = x & (1 << p) - 1;
int u = x & (1 << 20) - (1 << p);
return (u << 1) + v + d;
}
vector<int> Calc(int s,int t,int n) {
if(n == 1) {
vector<int> v;v.pb(s);v.pb(t);
return v;
}
int x = 1,p = 0;
while(1) {
if((s & x) != (t & x)) break;
x <<= 1;++p;
}
int a,b;
a = Delete(s,p);b = Delete(t,p);
int c = a ^ 1;
vector<int> f = Calc(a,c,n - 1),h = Calc(c,b,n - 1);
for(int i = 0 ; i < (1 << n - 1) ; ++i) {
f[i] = Insert(f[i],p,(s & x));
h[i] = Insert(h[i],p,(t & x));
}
f.insert(f.end(),h.begin(),h.end());
return f;
}
int count(int x) {
int res = 0;
while(x) {
res += (x & 1);
x >>= 1;
}
return res;
}
void Solve() {
read(N);read(A);read(B);
if(!(count(A^B) & 1)) {puts("NO");return;}
puts("YES");
vector<int> ans = Calc(A,B,N);
for(int i = 0 ; i < (1 << N) ; ++i) {
out(ans[i]);space;
}
enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
D - A Sequence of Permutations
設\(p\cdot q = t\)，\(t_{i} = p_{q_{i}}\)
這裏面\(a_{2} = q\cdot p^{-1}\)
咱們把這個寫成\(a = ABA^{-1}\)的形式
能夠獲得
設\(a = (A,B)\)
\(a_{1} = (id,p)\)
\(a_{2} = (id,q)\)
\(a_{3} = (id,qp^{-1})\)
\(a_{4} = (q,p^{-1})\)
\(a_5 = (qp^{-1},q^{-1})\)
\(a_6 = (qp^{-1},q^{-1}p)\)
\(a_{7} = (qp^{-1}q^{-1}p,p)\)
\(a_8 = (qp^{-1}q^{-1}p,q)\)
發現6次B一個重複
前面是\(qp^{-1}q^{-1}p\)的\(\lfloor \frac{K - 1}{6}\rfloor\)次冪
#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 MAXN 100005
#define eps 1e-10
//#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,K;
struct pl {
int f[100005];
friend pl operator * (const pl &a,const pl &b) {
pl c;
for(int i = 1 ; i <= N ; ++i) c.f[i] = a.f[b.f[i]];
return c;
}
pl inv() {
pl c;
for(int i = 1 ; i <= N ; ++i) {
c.f[f[i]] = i;
}
return c;
}
}p,q,iq,ip,s,res,t,ans;
void fpow(int c) {
for(int i = 1 ; i <= N ; ++i) res.f[i] = i;
t = s;
while(c) {
if(c & 1) res = res * t;
t = t * t;
c >>= 1;
}
}
void Solve() {
read(N);read(K);
for(int i = 1 ; i <= N ; ++i) read(p.f[i]);
for(int i = 1 ; i <= N ; ++i) read(q.f[i]);
ip = p.inv();iq = q.inv();
s = q * ip * iq * p;
fpow((K - 1) / 6);
int x = (K - 1) % 6;
if(x == 0) {
ans = res * p * res.inv();
}
else if(x == 1) {
ans = res * q * res.inv();
}
else if(x == 2) {
ans = res * q * ip * res.inv();
}
else if(x == 3) {
res = res * q;
ans = res * ip * res.inv();
}
else if(x == 4) {
res = res * q * ip;
ans = res * iq * res.inv();
}
else if(x == 5) {
res = res * q * ip;
ans = res * iq * p * res.inv();
}
for(int i = 1 ; i <= N ; ++i) {
out(ans.f[i]);space;
}
enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - Snuke the Phantom Thief
假如只能選K個，那麼咱們能夠建出限制
對於一維的，若是是小於等於\(a\)最多選\(b\)個，那麼至關於第\(b+1\)個必須在大於等於\(a + 1\)
若是大於等於\(a\)最多選\(b\)個，那麼至關於第\(K - b\)個必須在小於等於\(a - 1\)處
這樣能夠對於每個求一個左右限制出來
知足
\(L_{1} \leq L_{2} \leq L_{3}\leq \cdots \leq L_{k}\)
\(R_{1} \leq R_{2} \leq R_{3} \leq \cdots \leq R_{k}\)
而後對兩維都求這個限制
建K個點表示x軸上選的第幾個，K個點表示y軸上選的第幾個
而後2N個點表示每一個珠寶各兩個點，中間連一條容量爲1，價值爲珠寶價值的邊
而後前K個表示x軸上的點，和x符合條件的珠寶連一條容量爲1代價爲0的邊
而後每一個珠寶的第二個點和自身y匹配哪一個限制連一條容量爲1代價爲0的邊
而後原點向前K個連一條邊，後K個向匯點連一條邊，都是容量爲1價值爲0
#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 MAXN 100005
#define eps 1e-10
//#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,cap;int64 val;
}E[1000005];
int sumE,head[505],a[330],b[330],x[85],y[85],N,M;
int L[2][85],R[2][85],S,T,f;
int64 val[85],res,ans,ex;
bool vis[505];
char s[330][2];
void add(int u,int v,int c,int64 a) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].cap = c;
E[sumE].val = a;
head[u] = sumE;
}
void addtwo(int u,int v,int c,int64 a) {
add(u,v,c,a);
add(v,u,0,-a);
}
void build(int K) {
sumE = 1;memset(head,0,sizeof(head));
for(int i = 1 ; i <= N ; ++i) L[0][i] = L[1][i] = 1,R[0][i] = R[1][i] = 100;
for(int i = 1 ; i <= M ; ++i) {
if(b[i] >= K) continue;
if(s[i][0] == 'L') {
L[0][b[i] + 1] = max(L[0][b[i] + 1],a[i] + 1);
}
else if(s[i][0] == 'R') {
R[0][K - b[i]] = min(R[0][K - b[i]],a[i] - 1);
}
else if(s[i][0] == 'D') {
L[1][b[i] + 1] = max(L[1][b[i] + 1],a[i] + 1);
}
else if(s[i][0] == 'U') {
R[1][K - b[i]] = min(R[1][K - b[i]],a[i] - 1);
}
}
for(int i = 2 ; i <= K ; ++i) {
L[0][i] = max(L[0][i - 1],L[0][i]);
L[1][i] = max(L[1][i - 1],L[1][i]);
}
for(int i = K - 1 ; i >= 1 ; --i) {
R[0][i] = min(R[0][i + 1],R[0][i]);
R[1][i] = min(R[1][i + 1],R[1][i]);
}
S = 1;T = 2 * K + 2 * N + 2;
for(int i = 1 ; i <= K ; ++i) {
addtwo(S,i + 1,1,0);
addtwo(i + K + 2 * N + 1,T,1,0);
for(int j = 1 ; j <= N ; ++j) {
if(x[j] >= L[0][i] && x[j] <= R[0][i]) {
addtwo(i + 1,K + j + 1,1,0);
}
if(y[j] >= L[1][i] && y[j] <= R[1][i]) {
addtwo(K + N + j + 1,i + K + 2 * N + 1,1,0);
}
}
}
for(int i = 1 ; i <= N ; ++i) {
addtwo(K + i + 1,K + N + i + 1,1,val[i]);
}
}
int maxflow(int u,int aug) {
if(u == T) {
res += ex * aug;
f += aug;
return aug;
}
int flow = 0;
vis[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(E[i].cap > 0 && !vis[v] && E[i].val == 0) {
int t = maxflow(v,min(E[i].cap,aug - flow));
flow += t;
E[i].cap -= t;
E[i ^ 1].cap += t;
}
}
return flow;
}
bool modlabel() {
int64 p = -1e18;
for(int u = S ; u <= T ; ++u) {
if(vis[u]) {
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v] && E[i].cap > 0 && E[i].val > p) p = E[i].val;
}
}
}
if(p == -1e18) return false;
ex += p;
for(int u = S ; u <= T ; ++u) {
if(vis[u]) {
for(int i = head[u] ; i ; i = E[i].next) {
E[i].val -= p;
E[i ^ 1].val += p;
}
}
}
return true;
}
void Init() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(x[i]);read(y[i]);read(val[i]);
}
read(M);
for(int i = 1 ; i <= M ; ++i) {
scanf("%s",s[i]);read(a[i]);read(b[i]);
}
}
void Solve() {
ans = 0;
for(int i = 1 ; i <= N ; ++i) {
build(i);
res = 0;ex = 0;
f = 0;
do {
do {
memset(vis,0,sizeof(vis));
}while(maxflow(1,i));
}while(modlabel());
if(f == i) ans = max(ans,res);
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}
F - Walk on Graph
這道題特別神仙
反着來考慮，咱們設\((v,x)\)表示\(v\)這個點有價值爲\(x\)，那麼若是有一條邊\((u,v,c)\)，那麼咱們能夠走到的狀態是\((u,2x +c) \rightarrow (v,4x + 3c) \rightarrow (u,8x + 7c)\cdots\)
由於模數是奇數嘛，這麼乘下去，必定會獲得\(1\)的，感性理解可參照費馬小定理
那麼咱們就知道了
\((v,x)\rightarrow(u,2x + c)\)
\((u,2x+c)\rightarrow(v,x)\)
這是一個雙射
而後就是，若是一個點有代價爲\(a\)和代價爲\(b\)的邊直接相連，那麼它能夠走到
\((v,x)\Leftrightarrow (v,4x + 3a)\)
\((v,x)\Leftrightarrow (v,4x + 3a)\)
咱們能夠選擇這條路
\((v,4x + 3a) \rightarrow (v,x) \rightarrow(v,4x + 3b)\)
事實上，咱們存在對4的逆元，那麼\(4x\)能夠表示任何數，那麼咱們就又獲得了一條性質
\((v,x) \rightarrow (v,x + 3k(a - b)),k \in \Z\)
若是咱們對於全部的邊權之差和MOD求一個gcd，設爲\(g\)，\(g|a - b\)且\(g|b - c\)那麼\(g|a - c\)因此這個\(g\)能夠經過每條邊減第一條邊的絕對值的和MOD求gcd得到
事實上，咱們的運算能夠在\(gcd(MOD,3g)\)意義下進行
爲何呢，顯然咱們的模數要麼是\(g\)要麼是\(3g\)
若是是\(g\)的話，\(MOD\)和\(3\)互質，且必定存在兩個不一樣的邊權之差\(a - b\)和\(d - c\)互質，圖是聯通的，咱們能夠經過這兩個互質的邊權之差構造出，根據同餘方程，而後這個時候和3又互質，因此3存在逆元，咱們又能夠構造出任意數，因此模\(g\)意義下相等的就都相等了
\(3g\)也是同理的。。。
好的，這個問題解決了，那麼咱們修改了模數後，全部的邊權都相等了
那麼如今有個問題，若是\((v,x)\rightarrow (u,2x + c)\)
如今咱們是\((v,x)\rightarrow (u,2x + z + pg),(0 \leq p < 3)\)
看起來問題沒什麼實質性解決
可是若是咱們認爲\(x' = x + z\)邊權減去\(z\)
從\(x \rightarrow 2x + z + pg\)
就是\(x' \rightarrow 2x' + pg = 2x + 2z + pg - z\)
這樣的話邊權都是g的倍數了，也知足\((v,x) \rightarrow (v,2x + c)\)
這樣看起來很棒，那麼就這樣吧
因而咱們的詢問能夠從\((t,z)\)到\((s,r + z)\)
那麼一個點能夠被表示成
由於\((v,x) \rightarrow (v,4x + 3c)\)，並且\(c\)又是\(g\)的倍數，那麼認爲\((v,x) \rightarrow(v,4x)\)
那麼就有\((v,2^{q}x + pg)\)，\((0\leq q \leq 1,0 \leq p \leq 2)\)
這樣圖裏只有6個點了，咱們能夠用並查集連一下
查詢的時候
查\((t,x)\)和\((s,2^{p}x + qg)\)是否連通
這個能夠獲得
\(2^{p}z + qg = r + z\)
那麼看一下\(r + z - qg\)是2的奇數次冪乘\(x\)仍是2的偶數次冪乘\(x\)
#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 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 maxn = 5e5 + 5,maxm = 1e6 + 5;;
int N,M,Q,MOD,f[maxn],t[maxn],v[maxn];
int fa[maxm * 6],z,g;
bool able[2][maxm];
int id(int u,int p,int q) {
return (u - 1) * 6 + p * 3 + q + 1;
}
int gcd(int a,int b) {
return b == 0 ? a : gcd(b,a % b);
}
int getfa(int u) {
return u == fa[u] ? u : fa[u] = getfa(fa[u]);
}
void Merge(int u,int v) {
fa[getfa(u)] = getfa(v);
}
void Init() {
read(N);read(M);read(Q);read(MOD);
read(f[1]);read(t[1]);read(v[1]);
g = MOD;
for(int i = 2 ; i <= M ; ++i) {
read(f[i]);read(t[i]);read(v[i]);
g = gcd(g,abs(v[i] - v[1]));
}
MOD = gcd(MOD,3 * g);
z = v[1] % g;
int T = id(N,1,2);
for(int i = 1 ; i <= T; ++i) fa[i] = i;
for(int i = 1 ; i <= M ; ++i) {
int w = (v[i] / g) % 3;
for(int p = 0 ; p < 2 ; ++p) {
for(int q = 0 ; q < 3 ; ++q) {
Merge(id(f[i],p,q),id(t[i],p ^ 1,(q * 2 + w) % 3));
Merge(id(t[i],p,q),id(f[i],p ^ 1,(q * 2 + w) % 3));
}
}
}
for(int i = 0 , j = z ; i < MOD ; ++i , j = j * 2 % MOD) able[i & 1][j] = 1;
}
void Solve() {
int s,t,r;
for(int i = 1 ; i <= Q ; ++i) {
read(s);read(t);read(r);
bool res = 0;
for(int p = 0 ; p < 2 ; ++p) {
for(int q = 0 ; q < 3 ; ++q) {
if(getfa(id(t,0,0)) == getfa(id(s,p,q))) {
if(able[p][(r + z + (3 - q) * g) % MOD]) res = 1;
}
}
}
if(res) puts("YES");
else puts("NO");
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}
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相關文章
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