记次CF吧 1题。。。B题。。因为循环的i没设成long long 却参与了运算 结果就悲剧了 一直交 一直挂 。。上题
A 水。。 第一次少了个空格还。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<stdlib.h> 5 #include<algorithm> 6 using namespace std; 7 #define LL long long 8 LL x,y; 9 int main() 10 { 11 int i,j,k,n,m; 12 cin>>x>>y; 13 if((x<0&&y>0)||(x>0&&y<0)) 14 { 15 LL b = y-x; 16 if(b<0) 17 cout<<"0"<<" "<<b<<" "<<-b<<" "<<"0"<<endl; 18 else 19 cout<<-b<<" "<<"0 "<<"0"<<" "<<b<<endl; 20 } 21 else 22 { 23 LL b = y+x; 24 if(b>0) 25 cout<<"0"<<" "<<b<<" "<<b<<" "<<"0"<<endl; 26 else 27 cout<<b<<" "<<"0"<<" "<<"0 "<<b<<endl; 28 } 29 return 0; 30 }
B题 大体画画 就出来了 求下和加加
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<cmath> 7 using namespace std; 8 #define eps 1e-6 9 int main() 10 { 11 long long m,r,i; 12 cin>>m>>r; 13 double s =0,s1=0; 14 for(i = 1 ; i <= m ; i++) 15 { 16 s = 0; 17 if(i>2) 18 s+=(i-2)*(i-1)*r+(i-2.0)*sqrt(2.0)*2*r+sqrt(2.0)*r+2*r; 19 else 20 s+=(i-1)*i*r+(i-1)*sqrt(2.0)*r; 21 if(m-i>=2) 22 s+=(m-i-1)*(m-i)*r+(m-i-1)*sqrt(2.0)*2*r+2*r+sqrt(2.0)*r+2*r; 23 else 24 s+=(m-i)*(m-i+1)*r+(m-i)*sqrt(2.0)*r+2*r; 25 s1+=s/m; 26 } 27 printf("%.10f ",s1/m); 28 return 0; 29 }
往后就没再看 一直在交B 7次WA啊啊 泪~~
补道C题
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #define N 100010 7 #define LL long long 8 using namespace std; 9 struct node 10 { 11 LL a; 12 int d[50],k; 13 }q[N]; 14 int f[50]; 15 void digit(int e) 16 { 17 int g=0; 18 LL y = q[e].a; 19 while(y) 20 { 21 int x = y%2; 22 g++; 23 q[e].d[g] = x; 24 y/=2; 25 } 26 q[e].k = g; 27 } 28 bool cmp(node a,node b) 29 { 30 return a.a<b.a; 31 } 32 int main() 33 { 34 int i,j,k,n,o,g; 35 cin>>n; 36 for(i = 1; i <= n ;i++) 37 { 38 scanf("%d",&q[i].a); 39 digit(i); 40 } 41 sort(q+1,q+n+1,cmp); 42 o = 0; 43 for(i = 32; i >= 0 ; i--) 44 { 45 memset(f,0,sizeof(f)); 46 for(j = n; j >= 1 ; j--) 47 { 48 if(q[j].k<i) 49 break; 50 if(q[j].d[i]!=0) 51 { 52 for(g = 1; g < q[j].k ; g++) 53 if(q[j].d[g]==0) 54 f[g] = 1; 55 } 56 } 57 for(g = 1 ; g < i ; g++) 58 if(!f[g]) break; 59 if(g==i) 60 { 61 o = g; 62 break; 63 } 64 } 65 int num = 0; 66 for(i = n ; i >= 1 ; i--) 67 if(q[i].d[o]==1) 68 num++; 69 cout<<num<<endl; 70 int w=0; 71 for(i = n ; i >= 1 ; i--) 72 if(q[i].d[o]==1) 73 { 74 if(w) 75 printf(" "); 76 w++; 77 cout<<q[i].a; 78 } 79 return 0; 80 }
补道D题
根据费马小定理神马的求逆元 然后高端的算组合数取模
若第一个是1 则最后为0 若全是0且为偶数-》1 否则-》0 组合起来就可以求解了
100。。。。
0100。。
0010。。
0001。。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define mod 1000000007 8 #define LL long long 9 LL ff[200010]; 10 LL fmod(LL a,LL k) 11 { 12 LL b = 1; 13 while(k) 14 { 15 if(k&1) 16 b = a*b%mod; 17 a = (a%mod)*(a%mod)%mod; 18 k/=2; 19 } 20 return b; 21 } 22 LL cn(int n,int m) 23 { 24 LL ans,a; 25 ans = ff[n]; 26 a = fmod((ff[n-m]*ff[m])%mod,mod-2); 27 return (ans*a)%mod; 28 } 29 int main() 30 { 31 int i,j,k,n,m,g; 32 while(cin>>n>>m>>g) 33 { 34 LL s1=0,s2=0; 35 if(n==0) 36 { 37 if(m==1&&g==1) 38 cout<<"1 "; 39 else if(m>1&&g==1) 40 cout<<"0 "; 41 else if(m==1&&g==0) 42 cout<<"0 "; 43 else 44 cout<<"1 "; 45 continue; 46 } 47 else if(m==0) 48 { 49 if(n%2==0) 50 k = 1; 51 else 52 k = 0; 53 if(k==g) 54 cout<<"1 "; 55 else 56 cout<<"0 "; 57 continue; 58 } 59 ff[0] = 1; 60 for(i = 1;i <= n+m;i ++) 61 { 62 ff[i] = (ff[i-1]*i)%mod; 63 } 64 s1 = cn(n+m,m); 65 for(i = 0 ;i <= n ; i+=2) 66 { 67 if(m+n-i-1>=m-1) 68 s2=(s2+cn(m+n-i-1,m-1))%mod; 69 } 70 if(m==1&&n%2!=0) 71 s2++; 72 if(m==1&&n%2==0) 73 s2--; 74 if(g==0) 75 cout<<s2<<endl; 76 else 77 { 78 if(s1-s2<0) 79 s1 = s1-s2+mod; 80 else 81 s1 = s1-s2; 82 cout<<s1<<endl; 83 } 84 } 85 return 0; 86 }