Codeforces Round #242 (Div. 2) <A-C>_html/css_WEB-ITnose

CF424 A. Squats

题目意思：

有n（n为偶数）个x和X，求最少的变换次数，使得X的个数为n/2，输出变换后的序列。

解题思路：

统计X的个数ans,和n/2比较，少了的话，需要把n/2-ans个x变成X，多了的话需要把ans-n/2个X变成x.（从前往后扫一遍就行了）。

代码：

//#include<cspreadsheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt=(n/2-ans))               {                   printf("%c",save[i]);                   continue;               }               if(save[i]=='x')               {                   cnt++;                   printf("X");               }               else                    printf("X");           }       }       else       {           int cnt=0;           for(int i=1;i  <br>  CF 424B. Megacity  <p></p>  <p>题目意思：</p>  <p>给一个中心城市的坐标(0,0)和人口s，n个周围城市，告诉n个城市的人口及位置坐标，求以中心城市为圆心的最小的半径，使得人口总数超过1000000-s.</p>  <p>解题思路：</p>  <p>先求出每个城市距离中心城市的距离，然后对距离从小到大排序，然后依次扫描，如果达到要求，就退出输出最小的半径。</p>  <p>代码：</p>  <p></p>  <pre name="code" class="sycode">//#include<cspreadsheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt=1000000)       {           printf("0\n");           continue;       }       int lef=1000000-s;       int i=1;       while(lef>0&&i0)       {           printf("-1\n");           continue;       }       printf("%lf\n",ans);   }   return 0;}</cmath></bitset></ctime></queue></list></stack></set></map></vector></algorithm></cstring></string.h></string></cstdlib></sstream></cstdio></cmath></iostream></cspreadsheet.h>

CF 424C. Magic Formulas

题目意思：

给定pi,求Q。

解题思路：

抑或运算满足交换律和结合律。

原式可以等价于先对pi全部抑或，然后对每个i（1=

预处理出dp[i]=1^2^3..^i

代码：

//#include<cspreadsheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt  <br>  <br>  <p></p>  <p><br> </p>  <p><br> </p> </cmath></bitset></ctime></queue></list></stack></set></map></vector></algorithm></cstring></string.h></string></cstdlib></sstream></cstdio></cmath></iostream></cspreadsheet.h>