• hdu 4622 **


    题意:Suppose there are the symbols M, I, and U which can be combined to produce strings of symbols called "words". We start with one word MI, and transform it to get a new word. In each step, we can use one of the following transformation rules:
    1. Double any string after the M (that is, change Mx, to Mxx). For example: MIU to MIUIU.
    2. Replace any III with a U. For example: MUIIIU to MUUU.
    3. Remove any UU. For example: MUUU to MU.
    Using these three rules is it possible to change MI into a given string in a finite number of steps?

    首先全部转化为I

    可以发现,能被6整除或者i的个数是奇数且不为1的不可以,其他都可以

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<cstring>
     5 #include<cmath>
     6 #include<queue>
     7 #include<map>
     8 using namespace std;
     9 #define MOD 1000000007
    10 const int INF=0x3f3f3f3f;
    11 const double eps=1e-5;
    12 typedef long long ll;
    13 #define cl(a) memset(a,0,sizeof(a))
    14 #define ts printf("*****
    ");
    15 const int MAXN=1005055;
    16 int n,m,tt;
    17 char s[MAXN];
    18 int main()
    19 {
    20     int i,j,k;
    21     #ifndef ONLINE_JUDGE
    22     freopen("1.in","r",stdin);
    23     #endif
    24     scanf("%d",&tt);
    25     while(tt--)
    26     {
    27         scanf("%s",s);
    28         if(s[0]!='M')
    29         {
    30             printf("No
    ");
    31             continue;
    32         }
    33         int cnt=0;
    34         bool flag=1;
    35         int len=strlen(s);
    36         for(i=1;i<len;i++)
    37         {
    38             if(s[i]=='M')
    39             {
    40                 flag=0;
    41                 break;
    42             }
    43             if(s[i]=='I')   cnt++;
    44             else cnt+=3;
    45         }
    46         if(((cnt%6==0)||cnt%2==1&&cnt!=1)||!flag)
    47         {
    48             printf("No
    ");
    49         }
    50         else
    51             printf("Yes
    ");
    52     }
    53 }
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  • 原文地址:https://www.cnblogs.com/cnblogs321114287/p/4621457.html
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