Codeforces Round #258 (Div. 2)
A
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 #include<list> 11 using namespace std; 12 #define N 135 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 19 20 int main() 21 { 22 23 int n,m; 24 cin>>n>>m; 25 if(min(n,m)%2) 26 puts("Akshat"); 27 else puts("Malvika"); 28 return 0; 29 }
B
排序后比较
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 #include<list> 11 using namespace std; 12 #define N 101000 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 int a[N],b[N]; 19 20 int main() 21 { 22 23 int n,i; 24 cin>>n; 25 for(i = 1; i <= n; i++) 26 { 27 cin>>a[i]; 28 b[i] = a[i]; 29 } 30 sort(a+1,a+n+1); 31 int f = 0; 32 int flag = 1,st = 1,en = 1; 33 for(i = 1; i <= n ;i++) 34 { 35 if(a[i]!=b[i]&&!f) 36 { 37 st = i; 38 f = 1; 39 } 40 else if(a[i]!=b[i]) 41 en = i; 42 } 43 int j = en; 44 for(i = st ; i <= en ; i++) 45 { 46 if(a[j--]!=b[i]) 47 { 48 flag = 0; 49 break; 50 } 51 } 52 if(flag) printf("yes\n%d %d\n",st,en); 53 else printf("no\n"); 54 return 0; 55 }
C
根据d1d2的符号 分四种情况讨论，注意k和n的取值范围
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 #include<list> 11 using namespace std; 12 #define N 101000 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 int a[N],b[N]; 19 20 int main() 21 { 22 23 int t,i,j; 24 cin>>t; 25 LL n,k,d1,d2; 26 while(t--) 27 { 28 LL tk; 29 scanf("%I64d%I64d%I64d%I64d",&n,&k,&d1,&d2); 30 31 tk = n-k; 32 if(n%3!=0) 33 { 34 puts("no"); 35 continue; 36 } 37 if(d1+d2+d2<=k&&3*(d1+d2)<=n) 38 { 39 LL k1 = d1+d2+d1; 40 if((tk>=k1&&(tk-k1)%3==0)) 41 { 42 puts("yes"); 43 continue; 44 } 45 } 46 if(max(d1,d2)+abs(d1-d2)<=k&&3*d1<=n&&3*d2<=n) 47 { 48 LL k2 = d1+d2; 49 if(tk>=k2&&(tk-k2)%3==0) 50 { 51 puts("yes"); 52 continue; 53 } 54 } 55 if((d1+d2+d1)<=k&&3*(d2+d1)<=n) 56 { 57 LL k3 = d2+d2+d1; 58 if((tk>=k3&&(tk-k3)%3==0)) 59 { 60 puts("yes"); 61 continue; 62 } 63 } 64 if((d1+d2)<=k&&3*max(d1,d2)<=n) 65 { 66 LL k4 = max(d1,d2)+abs(d2-d1); 67 if(tk>=k4&&(tk-k4)%3==0) 68 { 69 puts("yes"); 70 continue; 71 } 72 } 73 puts("no"); 74 } 75 return 0; 76 }
D
因为只含ab 所以只要头尾相同肯定就是回文串，然后就统计一下第奇偶个位置上a,b的数目计数就可以了。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100010 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 char s[N]; 18 LL num[2][2],ans[2]; 19 int main() 20 { 21 int n,i,j; 22 cin>>s; 23 int k = strlen(s); 24 for(i = 0 ;i < k ;i++) 25 { 26 j = s[i]-'a'; 27 num[i&1][j]++; 28 ans[0]+=num[(i+1)&1][j]; 29 ans[1]+=num[i&1][j]; 30 } 31 cout<<ans[0]<<" "<<ans[1]<<endl; 32 return 0; 33 }
E
组合数学的知识。
组合数学P113有同等类型题目讲解。
具有重复的组合，可以先求出无限制的T*集合 然后根据容斥原理去除超出限制的组合个数。
T* = C（n+m-1,n) = C(n+m-1,m-1)
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100010 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 #define mod 1000000007 18 LL a[22],inv[22]; 19 LL getInv(LL x) 20 { 21 LL ret = 1; 22 x %= mod; 23 for (int a = mod - 2; a; a /= 2, x = x * x % mod) 24 if (a % 2 == 1) 25 ret = ret * x % mod; 26 return ret; 27 } 28 29 int C(LL n, int m) 30 { 31 if (n < m) 32 return 0; 33 int ret = 1; 34 for (LL i = n; i > n - m; -- i) 35 ret = ret * (i % mod) % mod * inv[n - i + 1] % mod; 36 return ret; 37 } 38 int main() 39 { 40 int n,i,j; 41 LL s; 42 for(i = 0 ; i <= 20 ; i++) 43 inv[i] = getInv(i); 44 cin>>n>>s; 45 for(i = 0 ; i< n; i++) 46 cin>>a[i]; 47 LL ans = C(s+n-1,n-1);//元素个数无限制的S集 48 for(i = 1; i < (1<<n) ; i++) 49 { 50 LL ss = s; 51 int o = 0; 52 for(j =0 ; j< n ; j++) 53 if(i&(1<<j)) 54 { 55 ss-=(a[j]+1);//超过限制的 56 o++; 57 } 58 if(ss<0) continue; 59 if(o&1) ans-=C(ss+n-1,n-1); 60 else ans+=C(ss+n-1,n-1); 61 ans = (ans%mod+mod)%mod;//mod 62 } 63 cout<<ans<<endl; 64 return 0; 65 }
