【AtCoder】ARC060
ARC060
C - 高橋君とカード / Tak and Cards
每个数减去A，然后转移N次，每次选或不选，最后是和为0的时候的方案数，负数可以通过把所有数右移2500做到
#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);
}
int N,A;
int x[55];
int64 dp[2][55 * 55 * 2];
int V = 2505;
void Solve() {
read(N);read(A);
for(int i = 1 ; i <= N ; ++i) {
read(x[i]);
x[i] -= A;
}
int cur = 0;
dp[cur][V] = 1;
for(int i = 1 ; i <= N ; ++i) {
memset(dp[cur ^ 1],0,sizeof(dp[cur ^ 1]));
for(int j = 0 ; j <= 2 * V ; ++j) {
if(j + x[i] >= 0) {
dp[cur ^ 1][j + x[i]] += dp[cur][j];
}
dp[cur ^ 1][j] += dp[cur][j];
}
cur ^= 1;
}
out(dp[cur][V] - 1);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
D - 桁和 / Digit Sum
小于1e6的可以暴力，大于1e6的显然只有两维数
\(N = kb + r,S = k + r\)
必要条件是\(N - S = k(b - 1)\)
枚举\(N - S\)的约数然后求出b看是否合法即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define 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);
}
int64 N,S,b = 1e18;
bool check(int64 x) {
if(N / x >= x) return false;
int64 k = N / x,r = N % x;
return k + r == S;
}
void Solve() {
read(N);read(S);
if(N == S) b = N + 1;
if(N > S) {
for(int64 i = 1 ; i <= (N - S) / i ; ++i) {
if((N - S) % i == 0) {
int64 t = (N - S) / i + 1;
if(check(t)) b = min(t,b);
int64 a = (N - S) / i;
t = (N - S) / a + 1;
if(check(t)) b = min(t,b);
}
}
}
for(int64 i = 2 ; i <= 1000000 ; ++i) {
int64 x = N,cnt = 0;
while(x) {
cnt += x % i;
x /= i;
}
if(cnt == S) {b = min(b,i);break;}
}
if(b == 1e18) {puts("-1");}
else {out(b);enter;}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
E - 高橋君とホテル / Tak and Hotels
这个倍增一下就好\(st[i][j]\)表示从i走\(2^j\)天能到的宾馆是啥
每次查询是log的
#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);
}
int N,L,Q;
int x[MAXN],st[MAXN][20];
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(x[i]);
read(L);read(Q);
for(int i = 1 ; i <= N ; ++i) {
int t = upper_bound(x + 1,x + N + 1,x[i] + L) - x - 1;
st[i][0] = t;
}
for(int j = 1 ; j <= 19 ; ++j) {
for(int i = 1 ; i <= N ; ++i) {
st[i][j] = st[st[i][j - 1]][j - 1];
}
}
int a,b;
for(int i = 1 ; i <= Q ; ++i) {
read(a);read(b);
if(a > b) swap(a,b);
int p = a,ans = 0;
for(int j = 19 ; j >= 0 ; --j) {
if(st[p][j] < b) {
p = st[p][j];
ans += (1 << j);
}
}
++ans;
out(ans);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
F - 最良表現 / Best Representation
假如所有字母一样，答案是\(|s|\)，方案数是1
假如最小循环节长度是\(|s|\)答案是1，方案数是1
剩下情况最小答案都是2，因为一定存在一种断开某个循环子串的中间使得两边都不为循环串，当循环次数为2的时候这个成立，当循环次数为3以上时，如果一个循环节尝试断开的每个位置两边都是循环串，那么这个循环节一定可以变小，与事实不符
注意判断循环节的方式是
N % (N - next[N]) == 0 ? N - next[N] : N
也就是有些情况下前后缀即使重合了，也可能没有循环节
之后我们只需要枚举位置使得两边的最小循环节都是自身就好了，取模1000000007是不存在的
#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 500005
//#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);
}
char s[MAXN];
int N;
int nxt[MAXN],suf[MAXN];
bool vis[MAXN];
int gcd(int a,int b) {
return b == 0 ? a : gcd(b,a % b);
}
void Solve() {
scanf("%s",s + 1);
N = strlen(s + 1);
for(int i = 2 ; i <= N ; ++i) {
int p = nxt[i - 1];
while(p && s[p + 1] != s[i]) p = nxt[p];
if(s[p + 1] == s[i]) nxt[i] = p + 1;
else nxt[i] = 0;
}
if(nxt[N] == N - 1) {
out(N);enter;puts("1");
}
else if(nxt[N] == 0 || N % (N - nxt[N]) != 0) {
puts("1");puts("1");
}
else {
int cnt = 0,l = N - nxt[N];
suf[N] = N + 1;
for(int i = N - 1 ; i >= 1 ; --i) {
int p = suf[i + 1];
while(p <= N && s[p - 1] != s[i]) p = suf[p];
if(s[p - 1] == s[i]) suf[i] = p - 1;
else suf[i] = N + 1;
}
for(int i = N - 1 ; i > 1 ; --i) {
if(suf[i] - i < (N + 1 - i) && (N + 1 - i) % (suf[i] - i) == 0) vis[i - 1] = 1;
}
for(int i = 2 ; i <= N ; ++i) {
if(i - nxt[i] < i && i % (i - nxt[i]) == 0) vis[i] = 1;
}
for(int i = 1 ; i < N ; ++i) {
if(!vis[i]) ++cnt;
}
puts("2");out(cnt);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
