UVa 580 - Critical Mass
2015-02-16 14:03
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题目:在一个队列中只包含L和U,如果3个U连续出现就非法,求给定长度的非法串的数量。
分析:dp。求出合法的数量,取补即可。
设f(XY,n)为长度为0、结尾字符为XY的合法串个数,则有:
f(UU,i)= f(LU,i-1);
f(UL,i)= f(UU,i-1)+ f(LU,i-1);
f(LU,i)= f(UL,i-1)+ f(LL,i-1);
f(LL,i)= f(UL,i-1)+ f(LL,i-1);
然后,求补即可(2^n - f(UU,n)- f(UL,n)- f(LU,n)- f(LL,n))。
说明:数据很小(n<=30)。
分析:dp。求出合法的数量,取补即可。
设f(XY,n)为长度为0、结尾字符为XY的合法串个数,则有:
f(UU,i)= f(LU,i-1);
f(UL,i)= f(UU,i-1)+ f(LU,i-1);
f(LU,i)= f(UL,i-1)+ f(LL,i-1);
f(LL,i)= f(UL,i-1)+ f(LL,i-1);
然后,求补即可(2^n - f(UU,n)- f(UL,n)- f(LU,n)- f(LL,n))。
说明:数据很小(n<=30)。
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include <cmath> using namespace std; int f[31][8]; int main() { for (int i = 0 ; i < 4 ; ++ i) f[2][i] = 1; for (int i = 3 ; i < 31 ; ++ i) { f[i][0] = f[i-1][2]; f[i][1] = f[i-1][0]+f[i-1][2]; f[i][2] = f[i][3] = f[i-1][1]+f[i-1][3]; } int n; while (~scanf("%d",&n) && n) { if (n < 3) printf("0\n"); else { int ans = 1<<n; for (int i = 0 ; i < 4 ; ++ i) ans -= f [i]; printf("%d\n",ans); } } return 0; }
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