C. Balance

題目鏈接

http://codeforces.com/contest/17/problem/C

題面

Nick likes strings very much, he likes to rotate them, sort them, rearrange characters within a string... Once he wrote a random string of characters a, b, c on a piece of paper and began to perform the following operations:

to take two adjacent characters and replace the second character with the first one,
to take two adjacent characters and replace the first character with the second one
To understand these actions better, let's take a look at a string ?abc?. All of the following strings can be obtained by performing one of the described operations on ?abc?: ?bbc?, ?abb?, ?acc?. Let's denote the frequency of a character for each of the characters a, b and c as the number of occurrences of this character in the string. For example, for string ?abc?: |a| = 1, |b| = 1, |c| = 1, and for string ?bbc?: |a| = 0, |b| = 2, |c| = 1.

While performing the described operations, Nick sometimes got balanced strings. Let's say that a string is balanced, if the frequencies of each character differ by at most 1. That is ?-?1?≤?|a|?-?|b|?≤?1, ?-?1?≤?|a|?-?|c|?≤?1 и ?-?1?≤?|b|?-?|c|?≤?1.

Would you help Nick find the number of different balanced strings that can be obtained by performing the operations described above, perhaps multiple times, on the given string s. This number should be calculated modulo 51123987.

輸入

The first line contains integer n (1?≤?n?≤?150) — the length of the given string s. Next line contains the given string s. The initial string can be balanced as well, in this case it should be counted too. The given string s consists only of characters a, b and c.

輸出

Output the only number — the number of different balanced strings that can be obtained by performing the described operations, perhaps multiple times, on the given string s, modulo 51123987.

樣例輸入

4
abca

樣例輸出

7

題意

你可以使得一個(gè)元素變成他周?chē)脑氐念伾?#xff0c;可以改變無(wú)數(shù)次，現(xiàn)在給你一個(gè)串，問(wèn)你一共有多少種方案，使得a和b和c的個(gè)數(shù)相差不超過(guò)1

題解

dp[i][a][b][c]，表示考慮到第i個(gè)位置，當(dāng)前有a個(gè)a，b個(gè)b，c個(gè)c 的方案數(shù)

然后轉(zhuǎn)移就好了

維護(hù)一個(gè)next[i][3]表示下一個(gè)在哪兒。

雖然是4維dp，但是卻是150 50 50 50 的

代碼

#include<bits/stdc++.h> using namespace std; const int mod = 51123987; int dp[152][52][52][52],n,nxt[152][3]; string s; void add(int &a,int b){a = a+b;if(a>=mod)a%=mod; } int main() {scanf("%d",&n);cin>>s;for(int j=0;j<3;j++)nxt[n][j]=n;for(int i=n-1;i>=0;i--){for(int j=0;j<3;j++)nxt[i][j]=nxt[i+1][j];nxt[i][s[i]-'a']=i;}dp[0][0][0][0]=1;int ans = 0;for(int i=0;i<n;i++){for(int a=0;a*3<=n+2;a++){for(int b=0;b*3<=n+2;b++){for(int c=0;c*3<=n+2&&a+b+c<=n;c++){if(dp[i][a][b][c]){if(a+b+c==n&&abs(b-c)<=1&&abs(a-c)<=1&&abs(b-c)<=1)add(ans,dp[i][a][b][c]);add(dp[nxt[i][0]][a+1][b][c],dp[i][a][b][c]);add(dp[nxt[i][1]][a][b+1][c],dp[i][a][b][c]);add(dp[nxt[i][2]][a][b][c+1],dp[i][a][b][c]);}}}}}printf("%d\n",ans); }

轉(zhuǎn)載于:https://www.cnblogs.com/qscqesze/p/6173531.html

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