【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('
')
#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('
')
#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('
')
#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('
')
#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;
}