A - Century

简单的除以200200200向上取整

B - 200th ABC-200

翻译成程序，whilewhilewhile模拟即可

C - Ringo’s Favorite Numbers 2

差为200200200倍数，则两个数同余200200200，按照模完后的余数分类单独计算即可Cri2C_{r_i}^2Cri2

#include <cstdio>
#define maxn 200005
#define int long long
int n;
int A[maxn], r[205];signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ )scanf( "%lld", &A[i] ), r[A[i] % 200] ++;int ans = 0;for( int i = 0;i < 200;i ++ )ans += ( r[i] * ( r[i] - 1 ) / 2 );printf( "%lld\n", ans );return 0;
}

D - Happy Birthday! 2

设dpi,j:dp_{i,j}:dpi,j:前iii个数中选某些数的和模200200200余数为jjj（iii必选）的方案数

直接大力转移，顺便用维克托儿记录下转移路径，方案数一旦等于222即代表有解

因为DPDPDP定义限制，两个序列的最后一个一定是一样的，这可能导致无法记录下最后一位不一样的方案

所以在原序列后面加一个000，这样保证了两个数列的最后一位一定会是一样的

#include <cstdio>
#include <vector>
using namespace std;
#define int long long
#define maxn 205
vector < pair < int, int > > G[maxn][maxn];
int n;
int A[maxn], ans1[maxn], ans2[maxn];
int f[maxn][maxn];void print( int *ans, int &cnt, int p, int r ) {if( ! G[p][r].size() ) return;ans[++ cnt] = p;print( ans, cnt, G[p][r][0].first, G[p][r][0].second );
}signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ )scanf( "%lld", &A[i] );A[++ n] = 0;f[0][0] = 1;int pos, r; bool flag = 1;for( int i = 1;i <= n && flag;i ++ )for( int j = 0;j < i && flag;j ++ )for( int k = 0;k <= 200 && flag;k ++ )if( f[j][k] ) {f[i][( k + A[i] ) % 200] += f[j][k];G[i][( k + A[i] ) % 200].push_back( make_pair( j, k ) );if( f[i][( k + A[i] ) % 200] == 2 ) {pos = i, r = ( k + A[i] ) % 200;flag = 0;break;}}if( ! flag ) {printf( "Yes\n" );int cnt1 = 0, cnt2 = 0, t = 0;ans1[++ cnt1] = ans2[++ cnt2] = pos;print( ans1, cnt1, G[pos][r][0].first, G[pos][r][0].second );print( ans2, cnt2, G[pos][r][1].first, G[pos][r][1].second );if( pos == n ) t = 1;printf( "%lld", cnt1 - t );for( int i = cnt1;i > t;i -- )printf( " %lld", ans1[i] );printf( "\n%lld", cnt2 - t );for( int i = cnt2;i > t;i -- )printf( " %lld", ans2[i] );}elseprintf( "No\n" );return 0;
}

E - Patisserie ABC 2

设dpi,jdp_{i,j}dpi,j表示选到第iii个数和为jjj的方案数量

发现状态转移dpi,j→dpi+1,j+1,dpi+1,j+2...dpi+1,j+ndp_{i,j}\rightarrow dp_{i+1,j+1},dp_{i+1,j+2}...dp_{i+1,j+n}dpi,j→dpi+1,j+1,dpi+1,j+2...dpi+1,j+n是连续段转移，用差分即可

接着线性即可扫出总和，然后枚举美丽度，计算出味道的取值范围，进而确定味道，自然而然就确定了欢迎度

#include <cstdio>
#include <iostream>
using namespace std;
#define int long long
int dp[4][3000005];
int n, k;signed main() {scanf( "%lld %lld", &n, &k );dp[0][0] = 1;for( int i = 0;i < 3;i ++ ) {for( int j = 0;j <= i * n;j ++ ) {dp[i + 1][j + 1] += dp[i][j];dp[i + 1][j + n + 1] -= dp[i][j];}for( int j = 1;j <= ( i + 1 ) * n;j ++ )dp[i + 1][j] += dp[i + 1][j - 1];}int sum;for( int i = 3;i <= 3 * n;i ++ )if( dp[3][i] >= k ) {sum = i;break;}elsek -= dp[3][i];for( int beauty = 1;beauty <= n;beauty ++ ) {int minn = max( 1ll, sum - beauty - n );int maxx = min( n, sum - beauty - 1 );if( minn > maxx ) continue;if( k > maxx - minn + 1 ) k -= ( maxx - minn + 1 );else {int taste = minn + k - 1;return ! printf( "%lld %lld %lld\n", beauty, taste, sum - beauty - taste );}}return 0;
}

F - Minflip Summation

对于每种可能串TTT，设S={i∣Ti≠Ti+1},i∈[0,len)S=\{i|T_i≠T_{i+1}\},i∈[0,len)S={i∣Ti=Ti+1},i∈[0,len)，举个栗子T=010110，S={0,1,2,4}

改写每次对一段l,r进行操作
{l≠1α(l−1)r≠len−1α(r)\begin{cases} l≠1&&\alpha(l-1)\\ r≠len-1&&\alpha(r)\\ \end{cases}\\ {l=1r=len−1α(l−1)α(r)
α(i):\alpha(i):α(i):如果Ti≠Ti+1T_i≠T_{i+1}Ti=Ti+1，把iii从SSS中删掉，否则加入其中

最后答案形态则是S=∅S=\emptyS=∅

不难发现，每次最多会从SSS中删去两个数，所以最后的答案为⌈tot2⌉\lceil\frac{tot}{2}\rceil⌈2tot⌉(tot=∣S∣tot=|S|tot=∣S∣)

接下来考虑统计所有情况串中相邻两个字符不同的数量

cnt=2kq×(k×∑i=0len−1f(i,i+1)+(k−1)×flen,0)cnt=2^{kq}\times (k\times \sum_{i=0}^{len-1}f(i,i+1)+(k-1)\times f_{len,0})cnt=2kq×(k×∑i=0len−1f(i,i+1)+(k−1)×flen,0)
fi,j={12(Ti=?)∣(Tj=?)1(Ti≠Tj)0(Ti=Tj)f_{i,j}=\begin{cases} \frac{1}{2}&&&(T_i=?)|(T_{j}=?)\\ 1&&&(T_i≠T_j)\\ 0&&&(T_i=T_j) \end{cases} fi,j=⎩⎪⎨⎪⎧2110(Ti=?)∣(Tj=?)(Ti=Tj)(Ti=Tj)

简单讨论一下

Ti=Tj=0/1T_i=T_j=0/1Ti=Tj=0/1，那么对于2kq2^{kq}2kq种情况的字符串iii都不可能被算进SSS，贡献000

Ti=0/1,Tj=1/0,Ti≠TjT_i=0/1,T_j=1/0,T_i≠T_jTi=0/1,Tj=1/0,Ti=Tj，同理对于2kq2^{kq}2kq种情况的字符串iii都会被算进SSS，贡献111

只有一个是???，那么只有12\frac{1}{2}21的概率贡献111，与另一个确定的字符不一样

两个都是???，一共有444种情况，两种相同（都是111，都是000），概率同样是12\frac{1}{2}21

重复KKK次原串，那么除了首尾少一次，其余都要×k\times k×k

但是这里仍然存在有一点小问题。

10100有三对相邻字符不一样，101001有四对相邻字符不一样，但是都需要花费222次操作

我们本意是想上取整，但是在模意义下非常难搞

事实上，对于一个有奇数对相邻字符不一样的串，重复多次后，理想与实际操作之间是恒定的

举个例子，101100，三对不一样，花费222，实际算的是111，重复一遍101100101100，七对不一样，花费444，实际算的是333……

归纳一下，也就是说对于奇数对的串答案总是少了111

加上这奇数对串的个数即为正确结果

不难发现，奇数对串的个数恰恰为首尾字符不相等的个数，2kq×f(T0,Tlen−1)2^{kq}\times f(T_0,T_{len-1})2kq×f(T0,Tlen−1)

所以最后的答案是cnt=2kq×k×∑i=0len−1f(i,(i+1)%len))cnt=2^{kq}\times k\times \sum_{i=0}^{len-1}f(i,(i+1)\%len))cnt=2kq×k×∑i=0len−1f(i,(i+1)%len))

#include <cstdio>
#include <cstring>
#define int long long
#define mod 1000000007
#define maxn 100005
char s[maxn];
int k, inv, ans;int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}int f( char x, char y ) {if( x == '?' || y == '?' ) return inv;else if( x != y ) return 1;else return 0;
}signed main() {scanf( "%s %lld", s, &k );int len = strlen( s );if( len * k == 1 ) return ! printf( "0\n" );int tot = 0;inv = qkpow( 2, mod - 2 );for( int i = 0;i < len;i ++ )tot += ( s[i] == '?' );int t = qkpow( 2, k * tot ) * k % mod;for( int i = 0;i < len;i ++ )ans = ( ans + t * f( s[i], s[( i + 1 ) % len] ) % mod ) % mod;ans = ans * inv % mod;printf( "%lld\n", ans );return 0;
}

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