DIV1 250pt

题意：对于一个字符串s，若对于每一个i = 0 to s.size()-p-1都有s[i] = s[i+p]则称字符串s是p循环的。"CATCATC", "CATCAT", "ACTAC" , "ACT"都是3循环。给定字符串s和一个数字ma，问至少修改多少个s中字符才能使得s变为k循环的，且k <= ma。

解法：水题，暴力枚举k = 1 to ma即可。

tag:think

1 // BEGIN CUT HERE  2 /*  3  * Author:  plum rain  4  * score :  5  */  6 /*  7   8  */  9 // END CUT HERE 10 #line 11 "DNAString.cpp" 11 #include 
      
        12 #include 
       
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                           31 #include 
                          

VI; 49 typedef vector

VS; 50 typedef vector

VD; 51 typedef pair

pii; 52 typedef long long int64; 53 54 const double eps = 1e-8; 55 const double PI = atan(1.0)*4; 56 const int inf = 2139062143 / 2; 57 58 string s; 59 int d[10]; 60 string temp = "ACTG"; 61 62 int gao(int x) 63 { 64 int ret = 0; 65 for (int i = 0; i < x; ++ i){ 66 clr0 (d); 67 int j = i; 68 while (j < sz(s)){ 69 for (int t = 0; t < 4; ++ t) 70 if (temp[t] == s[j]) ++ d[t]; 71 j += x; 72 } 73 int ma = 0, sum = 0; 74 for (int k = 0; k < 4; ++ k) 75 ma = max(ma, d[k]), sum += d[k]; 76 ret += sum - ma; 77 } 78 return ret; 79 } 80 81 class DNAString 82 { 83 public: 84 int minChanges(int m, vector

dna){ 85 s = accumulate(dna.begin(), dna.end(), string("")); 86 int ans = inf; 87 for (int i = 1; i <= m; ++ i) ans = min(ans, gao(i)); 88 return ans; 89 } 90 91 // BEGIN CUT HERE 92 public: 93 void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); } 94 private: 95 template

string print_array(const vector

&V) { ostringstream os; os << "{ "; for (typename vector

::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); } 96 void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } } 97 void test_case_0() { int Arg0 = 3; string Arr1[] = { "ATAGATA"}; vector

Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 1; verify_case(0, Arg2, minChanges(Arg0, Arg1)); } 98 void test_case_1() { int Arg0 = 2; string Arr1[] = { "ACGTGCA"}; vector

Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 3; verify_case(1, Arg2, minChanges(Arg0, Arg1)); } 99 void test_case_2() { int Arg0 = 13; string Arr1[] = { "ACGCTGACAGATA"}; vector

Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 0; verify_case(2, Arg2, minChanges(Arg0, Arg1)); }100 void test_case_3() { int Arg0 = 1; string Arr1[] = { "AAAATTTCCG"}; vector

Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 6; verify_case(3, Arg2, minChanges(Arg0, Arg1)); }101 void test_case_4() { int Arg0 = 12; string Arr1[] = { "ACGTATAGCATGACA","ACAGATATTATG","ACAGATGTAGCAGTA","ACCA","GAC"}; vector

Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 20; verify_case(4, Arg2, minChanges(Arg0, Arg1)); }102 103 // END CUT HERE104 105 };106 107 // BEGIN CUT HERE108 int main()109 {110 // freopen( "a.out" , "w" , stdout ); 111 DNAString ___test;112 ___test.run_test(-1);113 return 0;114 }115 // END CUT HERE