• hdu 5590 ZYB's Biology


    Problem Description
    After getting 600 scores in NOIP ZYB(ZJ−267) begins to work with biological questions.Now he give you a simple biological questions:
    he gives you a DNA sequence and a RNA sequence,then he asks you whether the DNA sequence and the RNA sequence are 
    matched.
    
    The DNA sequence is a string consisted of A,C,G,T;The RNA sequence is a string consisted of A,C,G,U.
    
    DNA sequence and RNA sequence are matched if and only if A matches U,T matches A,C matches G,G matches C on each position. 
     
    Input
    In the first line there is the testcase T.
    
    For each teatcase:
    
    In the first line there is one number N.
    
    In the next line there is a string of length N,describe the DNA sequence.
    
    In the third line there is a string of length N,describe the RNA sequence.
    
    1≤T≤10,1≤N≤100
     
    Output
    For each testcase,print YES or NO,describe whether the two arrays are matched.
     
    Sample Input
    2
    4
    ACGT
    UGCA
    4
    ACGT
    ACGU
     
    Sample Output
    YES 
    NO
     
    Source
     1 #pragma comment(linker, "/STACK:1024000000,1024000000")
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<cmath>
     6 #include<math.h>
     7 #include<algorithm>
     8 #include<queue>
     9 #include<set>
    10 #include<bitset>
    11 #include<map>
    12 #include<vector>
    13 #include<stdlib.h>
    14 #include <stack>
    15 using namespace std;
    16 #define PI acos(-1.0)
    17 #define max(a,b) (a) > (b) ? (a) : (b)
    18 #define min(a,b) (a) < (b) ? (a) : (b)
    19 #define ll long long
    20 #define eps 1e-10
    21 #define MOD 1000000007
    22 #define N 106
    23 #define inf 1e12
    24 int n;
    25 char s1[N];
    26 char s2[N];
    27 int main()
    28 {
    29    int t;
    30    scanf("%d",&t);
    31    while(t--){
    32       scanf("%d",&n);
    33       scanf("%s",s1);
    34       scanf("%s",s2);
    35       int flag=1;
    36       for(int i=0;i<n;i++){
    37          if(s1[i]=='A'){
    38             if(s2[i]!='U'){
    39                flag=0;
    40                break;
    41             }
    42          }
    43          else if(s1[i]=='T'){
    44             if(s2[i]!='A'){
    45                flag=0;
    46                break;
    47             }
    48          }
    49          else if(s1[i]=='C'){
    50             if(s2[i]!='G'){
    51                flag=0;
    52                break;
    53             }
    54          }
    55          else if(s1[i]=='G'){
    56             if(s2[i]!='C'){
    57                flag=0;
    58                break;
    59             }
    60          }
    61       }
    62       if(flag){
    63          printf("YES
    ");
    64       }else{
    65          printf("NO
    ");
    66       }
    67 
    68 
    69    }
    70     return 0;
    71 }
    View Code
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  • 原文地址:https://www.cnblogs.com/UniqueColor/p/5037016.html
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