【AtCoder】CODE FESTIVAL 2017 qual A
A - Snuke's favorite YAKINIKU
……
#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 40005
#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);
}
string s;
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
cin >> s;
if(s.substr(0,4) == "YAKI") puts("Yes");
else puts("No");
}
B - fLIP
枚举N有几个按了，M有几列按了
对于一个点，行和列只有一个按了，那么这个点是黑的
#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 MAXN 40005
#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,M,K;
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
read(N);read(M);read(K);
for(int i = 0 ; i <= N ; ++i) {
for(int j = 0 ; j <= M ; ++j) {
int res = i * (M - j) + (N - i) * j;
if(res == K) {puts("Yes");return 0;}
}
}
puts("No");
return 0;
}
C - Palindromic Matrix
从最外一圈开始推，我们输出需要几个4个一样的，2个一样的，可能还有需要1个单个的
然后用每个字母开始填4个一样的，再填2个一样的，再填1个单个的
如果没填完，那么就构造不出来
#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 40005
#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,M;
int num[30],cnt[10];
char s[105][105];
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
read(N);read(M);
for(int i = 1 ; i <= N ; ++i) {
scanf("%s",s[i] + 1);
}
for(int i = 1 ; i <= N ; ++i) {
for(int j = 1 ; j <= M ; ++j) {
num[s[i][j] - 'a']++;
}
}
int lr = 1,rr = N,lc = 1,rc = M;
while(lr <= rr && lc <= rc) {
int r = 0;
if(lr != rr) r += 2;
if(lc != rc) r += 2;
if(!r) r = 1;
cnt[r]++;
for(int i = 1 ; i <= N ; ++i) {
if(lr + i <= rr - i) {
int r = 0;
if(lr + i != rr - i) r += 2;
if(lc != rc) r += 2;
cnt[r]++;
}
else break;
}
for(int i = 1 ; i <= M ; ++i) {
if(lc + i <= rc - i) {
int r = 0;
if(lc + i != rc - i) r += 2;
if(rr != lr) r += 2;
cnt[r]++;
}
else break;
}
++lr;--rr;++lc;--rc;
}
for(int i = 0 ; i < 26 ; ++i) {
while(cnt[4] && num[i] >= 4) {num[i] -= 4;--cnt[4];}
while(cnt[2] && num[i] >= 2) {num[i] -= 2;--cnt[2];}
while(cnt[1] && num[i] >= 1) {num[i] -= 1;--cnt[1];}
}
if(cnt[1] || cnt[2] || cnt[4]) puts("No");
else puts("Yes");
return 0;
}
D - Four Coloring
把图旋转45度后，再放一个每块大小是d*d的方格，然后一行用RB间隔染色，下一行用GY间隔染色
#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 H,W,d;
bool vis[1005][1005];
string s = "RBGY";
void Solve() {
read(H);read(W);read(d);
for(int i = 1 ; i <= H ; ++i) {
for(int j = 1 ; j <= W ; ++j) {
int t = 0;
if((i + j - 1) / d & 1) t |= 2;
if((i - j + 500 - 1) / d & 1) t |= 1;
putchar(s[t]);
}
enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - Modern Painting
第一次肯定是一个竖直的完整画下来，或者一个水平的完整的画下来
用竖直的举例
就是枚举一个区间\([L,R]\)都是完整的竖直画下来，方案数是2的这个区间上下都有人的列个数次幂
然后算前后的那一块方案数
肯定是先横着一刀，再竖着一刀，再横着一刀
我们把横着的轮廓线画出来，再把一边的翻过去，发现是一个路径计数，但是要求对称轴必须有一个竖线
加入竖直有X个人，上有Y人，下有Z人，那么如果直接走过去，方案数是
\(\binom{X + Y + Z}{X}\)
那我们考虑一个序列，我们就走\(X - 1\)个竖直的，横向走到对称轴后，紧接着就添加一个竖直的
方案数就是
\(\binom{X + Y + Z - 1}{X - 1}\)就是方案数
#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 200500
#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 = 998244353;
int N,M;
char a[4][MAXN];
int fac[MAXN * 2],invfac[MAXN * 2],f[MAXN],b[MAXN],sum[4][MAXN],s[MAXN];
int pw[MAXN],ipw[MAXN],ans;
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;
}
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 update(int &x,int y) {
x = inc(x,y);
}
int C(int n,int m) {
if(n < m) return 0;
return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void CalcV() {
memset(f,0,sizeof(f));
memset(b,0,sizeof(b));
memset(s,0,sizeof(s));
int X = sum[0][N],Y,Z;
for(int i = 0 ; i <= M ; ++i) {
if(a[2][i + 1] == '0' && a[3][i + 1] == '0') continue;
Y = sum[2][i],Z = sum[3][i];
if(!X) {
if(Y == 0 && Z == 0) f[i] = 1;
else f[i] = 0;
}
else {
f[i] = C(X + Y + Z - 1,X - 1);
}
}
X = sum[1][N];
for(int i = M + 1 ; i >= 1 ; --i) {
if(a[2][i - 1] == '0' && a[3][i - 1] == '0') continue;
Y = sum[2][M] - sum[2][i - 1],Z = sum[3][M] - sum[3][i - 1];
if(!X) {
if(Y == 0 && Z == 0) b[i] = 1;
else b[i] = 0;
}
else {
b[i] = C(X + Y + Z - 1,X - 1);
}
}
for(int i = 1 ; i <= M ; ++i) s[i] = s[i - 1] + (a[2][i] == '1' && a[3][i] == '1');
for(int i = 0 ; i <= M ; ++i) {
f[i] = mul(f[i],ipw[s[i]]);
}
for(int i = M + 1 ; i >= 1 ; --i) {
b[i] = inc(b[i + 1],mul(b[i],pw[s[i - 1]]));
}
for(int i = 1 ; i <= M ; ++i) {
update(ans,mul(f[i - 1],b[i + 1]));
}
}
void CalcH() {
memset(f,0,sizeof(f));
memset(b,0,sizeof(b));
memset(s,0,sizeof(s));
int X = sum[2][M],Y,Z;
for(int i = 0 ; i <= N ; ++i) {
if(a[0][i + 1] == '0' && a[1][i + 1] == '0') continue;
Y = sum[0][i],Z = sum[1][i];
if(!X) {
if(Y == 0 && Z == 0) f[i] = 1;
else f[i] = 0;
}
else {
f[i] = C(X + Y + Z - 1,X - 1);
}
}
X = sum[3][M];
for(int i = N + 1 ; i >= 1 ; --i) {
if(a[0][i - 1] == '0' && a[1][i - 1] == '0') continue;
Y = sum[0][N] - sum[0][i - 1],Z = sum[1][N] - sum[1][i - 1];
if(!X) {
if(Y == 0 && Z == 0) b[i] = 1;
else b[i] = 0;
}
else {
b[i] = C(X + Y + Z - 1,X - 1);
}
}
for(int i = 1 ; i <= N ; ++i) s[i] = s[i - 1] + (a[0][i] == '1' && a[1][i] == '1');
for(int i = 0 ; i <= N ; ++i) {
f[i] = mul(f[i],ipw[s[i]]);
}
for(int i = N + 1 ; i >= 1 ; --i) {
b[i] = inc(b[i + 1],mul(b[i],pw[s[i - 1]]));
}
for(int i = 1 ; i <= N ; i++) {
update(ans,mul(f[i - 1],b[i + 1]));
}
}
void Solve() {
read(N);read(M);
for(int i = 0 ; i <= 3 ; ++i) scanf("%s",a[i] + 1);
for(int i = 0 ; i <= 3 ; ++i) {
int T = (i <= 1 ? N : M);
for(int j = 1 ; j <= T ; ++j) {
sum[i][j] = sum[i][j - 1] + (a[i][j] == '1');
}
}
int T = 2 * (N + M) + 10;
fac[0] = 1;
for(int i = 1 ; i <= T ; ++i) {
fac[i] = mul(fac[i - 1],i);
}
invfac[T] = fpow(fac[T],MOD - 2);
for(int i = T - 1 ; i >= 0 ; --i) {
invfac[i] = mul(invfac[i + 1],i + 1);
}
pw[0] = 1;
T = max(N,M);
for(int i = 1 ; i <= T ; ++i) {
pw[i] = mul(pw[i - 1],2);
}
ipw[0] = 1;
for(int i = 1 ; i <= T ; ++i) {
ipw[i] = mul(ipw[i - 1],(MOD + 1) / 2);
}
CalcV();
CalcH();
if(!sum[0][N] && !sum[1][N] && !sum[2][M] && !sum[3][M]) ans = 1;
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
F - Squeezing Slimes
对于一个单个的区间来说，我们可以用
如果\(a = 2^{k}\)能用k次合成
如果\(2^{k} < a < 2^{k + 1}\)，能用k+1次合成
我们记录一下每个区间最左的元素被选了几次，最右的区间选了几次
如果是\(a = 2^{k}\)，那么两边就选了k次
另一种有两个情况，左边k次右边k+1次，左边k+1次右边k次
每个相邻的地方可以减少\(min(r_{i},l_{i + 1})\)
用一个dp实现就行
#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-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;
int a[MAXN],cnt[MAXN],val[MAXN][2];
int64 dp[MAXN][2];
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(a[i]);
int k = -1,x = a[i];
while(x) {++k;x >>= 1;}
cnt[i] = k;
}
for(int i = 1 ; i <= N ; ++i) {
dp[i][0] = 1e16;dp[i][1] = 1e16;
int s = cnt[i],t = cnt[i];
if(a[i] != (1 << cnt[i])) ++t;
val[i][0] = s;val[i][1] = t;
for(int j = 0 ; j <= 1 ; ++j) {
dp[i][0] = min(dp[i][0],dp[i - 1][j] + t - min(t,val[i - 1][j]));
dp[i][1] = min(dp[i][1],dp[i - 1][j] + t - min(s,val[i - 1][j]));
}
}
out(min(dp[N][0],dp[N][1]));enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
