CF467 AB 水题

Codeforces Round #267 (Div. 2)

（C和D的题解单独写：CF467C George and Job （DP） CF467D Fedor and Essay 建图DFS）

CF#267A

CF#267B

A

George and Accommodation

题意：给出宿舍各个寝室住了多少人、最多能住多少人，求有多少个寝室能住进这两个人……

代码：

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usll unsigned ll
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)
 #define mp make_pair
 #define pb push_back
 const double eps=1e-;
 const double pi=acos(-1.0);

 int main(){
     int n,x,y,ans=;
     RD(n);
     while(n--){
         RD2(x,y);
         if(y-x>=)ans++;
     }
     WN(ans);
     return ;
 }

B

Fedor and New Game

题意：给出n,m,k，给出m+1个数a[i]，代表m+1个人的那个什么情况，主角是第m+1个人。求a[i]和a[m+1]的n个二进制位中，有不超过k位不同的i有多少个（1<=i<=m)。

题解：直接暴力搞啊

代码：

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usll unsigned ll
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)
 #define mp make_pair
 #define pb push_back
 const double eps=1e-;
 const double pi=acos(-1.0);
 int a[];
 int n,m,k;
 int main(){
     int i,j;
     RD3(n,m,k);
     REP(i,m+){
         RD(a[i]);
     }
     int q=a[m];
     int ans=;
     REP(i,m){
         int t=q^a[i];
         int dif=;
         REP(j,n){
             if((t&)==)dif++;
             t>>=;
         }
         if(dif<=k)ans++;
     }
     WN(ans);
     return ;
 }