CF467 AB 水题

（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

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 ;

 }

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 ;

 }

巴特西

CF467 AB 水题

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