codeforces AIM Tech Round

感觉半夜撸代码就像傻逼。。

Graph and String

用a和c二分染色，

#include <bits/stdc++.h>

using namespace std;


#define N 1000
#define M 2000000
#define ULL unsigned long long 
#define LL long long
#define mod 1000000007


int fst[N], vv[M], nxt[M], e;
void init() {
e = 0;
memset(fst, -1, sizeof fst);
}


void add(int u,int v){ 
vv[e] = v, nxt[e] = fst[u], fst[u] = e ++;
}
int g[N][N];
bool vis[N];
bool flag;
char s[N];


void dfs(int u, int p) {
vis[u] = 1;
s[u] = p+'a';
p = 2-p;
for(int i = fst[u]; ~i; i = nxt[i]) {
int v = vv[i];
if(vis[v]) {
if(s[v] != p+'a')
flag = 1;
continue;
}
dfs(v, p);
}
}




int main() {
int n, m;
scanf("%d%d", &n, &m);
init();
int u, v;
for(int i = 0; i < m; ++i) 
scanf("%d%d", &u, &v), g[u][v] = g[v][u] = 1;
for(int i = 1; i <= n; ++i) 
for(int j = i+1; j <= n; ++j) 
if(g[i][j] == 0)
add(i, j), add(j, i);
for(int i = 1; i <= n; ++i) if(!vis[i]){
if(fst[i] == -1) s[i] = 'b';
else dfs(i, 0);
}
if(flag) {
puts("No");
return 0;
}
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) 
{
if(i == j) continue;
if(g[i][j] == 1 && abs(s[i]-s[j]) > 1)
{

puts("No");
return 0;
}
if(g[i][j] == 0 && abs(s[i] - s[j]) <= 1)
{
puts("No");
return 0;
}
}
}
s[n+1] = 0;
printf("Yes\n%s\n", s+1);

}

Array GCD

题意：

操作一 最多删掉一个区间，花费len×a 

操作二 a[i] +- 1, 花费b

把序列gcd改为不为1的最小花费

由于a[1] 或a[n] 不能删

枚举他们+-1的质因子，它的倍数作为最后的gcd 的

dp[i][0]表示还没有开始删区间

dp[i][1]表示正在删区间

dp[i][2]表示已经删完了一个区间

#include <bits/stdc++.h>using namespace std;#define N (1000000 + 10)#define M 2000000#define inf (long long)1e17#define ULL unsigned long long #define LL long long#define mod 1000000007int a,b, c[N];int cnt, d[N];int n;void gao(int x) {for(int i = 2; i * i <= x; ++i) {if(x % i == 0) {d[cnt++] = i;while(x % i == 0) x /= i;}}if(x > 1) d[cnt++] = x;}LL dp[N][3];LL f(int p) {dp[0][0] = 0;dp[0][1] = dp[0][2] = inf;for(int i = 1; i <= n; ++i) {for(int j = 0; j < 3; ++j)dp[i][j] = inf;if(c[i] % p == 0) {dp[i][0] = dp[i-1][0];dp[i][1] = min(dp[i-1][0] + a, dp[i-1][1] + a);dp[i][2] = min(dp[i-1][2], dp[i-1][1]);}else if((c[i]+1)%p == 0 || (c[i]-1+p)%p == 0) {dp[i][0] = dp[i-1][0] + b;dp[i][1] = min(dp[i-1][0] + a, dp[i-1][1] + a);dp[i][2] = min(dp[i-1][2] + b, dp[i-1][1] + b);}else {dp[i][1] = min(dp[i-1][0] + a, dp[i-1][1] + a);}}return min(min(dp[n][0], dp[n][1]), dp[n][2]);}int main() {scanf("%d%d%d", &n, &a, &b);for(int i = 1; i <= n; ++i)scanf("%d", &c[i]);for(int i = -1; i <= 1; ++i) {gao(c[1]+i); gao(c[n] + i);}sort(d, d + cnt);cnt = unique(d, d+cnt) - d;LL ans = inf;for(int i = 0; i < cnt; ++i)ans = min(ans, f(d[i]));printf("%I64d\n", ans);}

Birthday

奇葩概率题，好像暴力算期望

Transforming Sequence

题意：

用1到2^k-1的数组成a[1] 到a[n]

使得b[i]严格递增的方案数有多少，b[i] = a[1] | a[2]……| a[i]

dp[n][k]表示前n个数二进制位上共k个1

转移是n^3

用倍增省掉第一维然后用fft加速 做到nlognlogn

涨姿势啊

#include <bits/stdc++.h>using namespace std;#define pii pair<int, int>#define MP make_pair#define ls (i<<1)#define rs (i<<1|1)#define md (ll+rr>>1)#define N (1000000 + 10)#define M (2000000 + 10)#define inf 0x3f3f3f3f#define ULL unsigned long long #define LL long long#define mod 1000000007// 多项式乘法 系数对MOD=1000000007取模， 常数巨大，慎用// 只要选的K个素数乘积大于MOD*MOD*N,理论上MOD可以任取。#define MOD 1000000007#define K 3const int m[K] = {1004535809, 998244353, 104857601};#define G 3int qpow(int x, int k, int P) {int ret = 1;while(k) {if(k & 1) ret = 1LL * ret * x % P;k >>= 1;x = 1LL * x * x % P;}return ret;}struct _NTT {int wn[25], P;void init(int _P) {P = _P;for(int i = 1; i <= 21; ++i) {      int t = 1 << i;      wn[i] = qpow(G, (P - 1) / t, P);      }    }void change(int *y, int len) {for(int i = 1, j = len / 2; i < len - 1; ++i) {      if(i < j) swap(y[i], y[j]);      int k = len / 2;      while(j >= k) {      j -= k;      k /= 2;      }      j += k;      } }void NTT(int *y, int len, int on) {change(y, len);      int id = 0;            for(int h = 2; h <= len; h <<= 1) {      ++id;      for(int j = 0; j < len; j += h) {      int w = 1;      for(int k = j; k < j + h / 2; ++k) {      int u = y[k];      int t = 1LL * y[k+h/2] * w % P;     y[k] = u + t;      if(y[k] >= P) y[k] -= P;      y[k+h/2] = u - t + P;      if(y[k+h/2] >= P) y[k+h/2] -= P;  w = 1LL * w * wn[id] % P;}      }      }      if(on == -1) {      for(int i = 1; i < len / 2; ++i) swap(y[i], y[len-i]);      int inv = qpow(len, P - 2, P);      for(int i = 0; i < len; ++i)   y[i] = 1LL * y[i] * inv % P;}      }void mul(int A[], int B[], int len) {NTT(A, len, 1);NTT(B, len, 1);for(int i = 0; i < len; ++i) A[i] = 1LL * A[i] * B[i] % P;NTT(A, len, -1);}}ntt[K];int tmp[N][K], t1[N], t2[N];int r[K][K];int CRT(int a[]) {int x[K];for(int i = 0; i < K; ++i) {x[i] = a[i];for(int j = 0; j < i; ++j) {int t = (x[i] - x[j]) % m[i];if(t < 0) t += m[i];x[i] = 1LL * t * r[j][i] % m[i];}}int mul = 1, ret = x[0] % MOD;for(int i = 1; i < K; ++i) {mul = 1LL * mul * m[i-1] % MOD;ret += 1LL * x[i] * mul % MOD;if(ret >= MOD) ret -= MOD;}return ret;}void mul(int A[], int B[], int len) {for(int id = 0; id < K; ++id) {for(int i = 0; i < len; ++i) {t1[i] = A[i];t2[i] = B[i];}ntt[id].mul(t1, t2, len);for(int i = 0; i < len; ++i) tmp[i][id] = t1[i];}for(int i = 0; i < len; ++i) {A[i] = CRT(tmp[i]);}}void init() {for(int i = 0; i < K; ++i) {for(int j = 0; j < i; ++j) {r[j][i] = qpow(m[j], m[i] - 2, m[i]);}}for(int i = 0; i < K; ++i) {ntt[i].init(m[i]);}}int ff[N], nff[N];int dp[N];int ans[N];int pow2[N];int ta[N], tb[N];void mymul(int *A, int *B, int k, int lb) {int tlen = 1;while(tlen < (k+1) * 2) tlen <<= 1;for(int i = 0; i <= tlen; ++i)ta[i] = tb[i] = 0;for(int i = 0; i <= k; ++i)ta[i] = 1LL * A[i] * nff[i] % mod * qpow(2, i*lb, mod) % mod;for(int i = 0; i <= k; ++i)tb[i] = 1LL * B[i] * nff[i] % mod;//debug//for(int i = 0; i <= k; ++i)//printf("aa %d bb %d\n", ta[i], tb[i]);//mul(ta, tb, tlen);//debug//for(int i = 0; i <= k; ++i)//printf("ta %d\n", ta[i]);for(int i = 0; i <= k; ++i)A[i] = 1LL * ta[i] * ff[i] % mod;}int main() {LL nn;int n, k;scanf("%I64d%d", &nn, &k);if(nn > k) {puts("0");return 0;}n = nn;ff[0] = nff[0] = 1;for(int i = 1; i <= k*30; ++i)ff[i] = 1LL * ff[i-1] * i % mod, nff[i] = qpow(ff[i], mod-2, mod);pow2[0] = 1;for(int i = 1; i <= k*30; ++i)pow2[i] = 1LL * pow2[i-1] * 2 %mod;for(int i = 1; i <= k; ++i) dp[i] = 1;init();int len = 1;ans[0] = 1;for(int z = n; z; z >>= 1) {if(z & 1) {mymul(ans, dp, k, len);//for(int i = 0; i <= k; ++i)//printf("&1 r %d\n", ans[i]);}mymul(dp, dp, k, len);//for(int i = 0; i <= k; ++i)//printf("dp %d\n", dp[i]);len += len;}int res = 0;for(int i = 0; i <= k; ++i)res = (res + 1LL * ans[i] * ff[k] % mod * nff[i] % mod * nff[k-i] % mod) %mod;printf("%d\n", res);}