Codeforces Round #242 (Div. 2) _html/css_WEB-ITnose

陆砚码

2026-10-12 • 👁 0人浏览过

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）。立即学习 “ 前端免费学习笔记（深入） ”；代码：

//#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;char save[220];int n;int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   while(~scanf("%d",&n))   {       int ans=0;       scanf("%s",save+1);       for(int i=1;i<=n;i++)            if(save[i]=='X')                ans++;       printf("%d\n",abs(n/2-ans));       if(ans=(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<=n;i++)           {               if(cnt==(ans-n/2))               {                   printf("%c",save[i]);                   continue;               }               if(save[i]=='X')               {                   cnt++;                   printf("x");               }               else                    printf("x");           }       }       putchar('\n');   }   return 0;}

CF 424B. Megacity 题目意思：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. 解题思路：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。代码：

//#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 1100struct Point{    double x,y;    double dis;    int v;}pp[Maxn];int n,s;int dp[Maxn];bool cmp(struct Point a,struct Point b){    return a.dis=1000000)       {           printf("0\n");           continue;       }       int lef=1000000-s;       int i=1;       while(lef>0&&i<=n)       {           ans=pp[i].dis;           lef-=pp[i].v;           i++;       }       if(lef>0)       {           printf("-1\n");           continue;       }       printf("%lf\n",ans);   }   return 0;}

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#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 1100000int dp[Maxn];int n;void init(){    dp[0]=0;    for(int i=1;i<=1000000;i++)        dp[i]=dp[i-1]^i;}int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   init();   while(~scanf("%d",&n))   {       int ans=0;       for(int i=1;i<=n;i++)       {           int p;           scanf("%d",&p);           ans=ans^p;           if((n/i)&1) //ÆæÊý           {               ans=ans^dp[i-1];           }           ans=ans^dp[n%i];       }       printf("%d\n",ans);   }   return 0;}

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