A. Splits
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <vector> 7 #include <set> 8 #include <list> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 using namespace std; 13 const long maxn=1e5; 14 const long mod=1e9+7; 15 16 17 int main() 18 { 19 long n,i; 20 scanf("%ld",&n); 21 printf("%ld",1+n/2); 22 return 0; 23 }
B. Messages
贪心
#include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <vector> #include <set> #include <list> #include <map> #include <queue> #include <iostream> #include <algorithm> using namespace std; const long maxn=1e5; const long mod=1e9+7; int main() { long n,a,b,c,T,t,v,i,sum=0; scanf("%ld%ld%ld%ld%ld",&n,&a,&b,&c,&T); sum=a*n; for (i=1;i<=n;i++) { scanf("%ld",&t); if (b<=c) sum=sum+(c-b)*(T-t); } cout<<sum<<endl; return 0; }
C. Alternating Sum
快速幂 求逆元 等比数列
1. 1e9+9不是素数
2. 等比数列，公比可能为0
3. 数的数目为n+1
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <vector> 7 #include <set> 8 #include <list> 9 #include <map> 10 #include <queue> 11 #include <iostream> 12 #include <algorithm> 13 using namespace std; 14 const long maxn=1e5; 15 const long long mod=1e9+9; 16 #define ll long long 17 18 ll x,y; 19 20 ll mul(ll s,long ci) 21 { 22 ll r=1; 23 while (ci) 24 { 25 if ((ci & 1)==1) 26 r=r*s%mod; 27 s=s*s%mod; 28 ci=ci>>1; 29 } 30 return r; 31 } 32 33 void gcd(ll a,ll b) 34 { 35 if (b==0) 36 { 37 x=1; 38 y=0; 39 } 40 else 41 { 42 ll r; 43 gcd(b,a%b); 44 r=x; 45 x=y; 46 y=r-a/b*y; 47 } 48 } 49 50 ll ni(ll s) 51 { 52 gcd(mod,s); 53 return (y%mod+mod)%mod; 54 } 55 56 int main() 57 { 58 long n,k,i; 59 long long a,b,ni_a,ni_a_k,a_k,b_k,sum,c,d; 60 char ch; 61 scanf("%ld%lld%lld%ld\n",&n,&a,&b,&k); 62 c=mul(a,n); 63 64 b_k=mul(b,k); 65 a_k=mul(a,k); 66 67 ni_a=ni(a); 68 ni_a_k=ni(a_k); 69 70 sum=0; 71 for (i=1;i<=k;i++) 72 { 73 scanf("%c",&ch); 74 if (ch=='+') 75 sum=(sum+c)%mod; 76 else 77 sum=(sum-c+mod)%mod; 78 } 79 80 d=(n+1)/k; 81 c=ni_a_k * b_k %mod; 82 83 if (c==1) 84 printf("%lld",sum*d%mod); 85 else 86 printf("%lld",sum* ((mul(c,d)%mod-1+mod)%mod) %mod *ni(c-1)%mod); 87 88 return 0; 89 }
D. Destruction of a Tree
树结构，从叶子向根，先确定子节点是否删边，然后父节点是否删边就可以确定，当根节点不删边，则成立，否则不成立
输出删点次序：从叶子向根，遇到需要删边的点，则删边，通过遍历所有的子辈节点，遇到不需要删除边的点，则输出，继续遍历，否则结束
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <vector> 7 #include <set> 8 #include <list> 9 #include <map> 10 #include <queue> 11 #include <iostream> 12 #include <algorithm> 13 using namespace std; 14 const long maxn=2e5+100; 15 const long long mod=1e9+9; 16 #define ll long long 17 18 struct node 19 { 20 long d; 21 struct node *next; 22 }*point[maxn]; 23 long need[maxn],fa[maxn]; 24 bool vis[maxn],feng[maxn]; 25 long pr[maxn],pr_count=0; 26 27 28 void dfs(long d) 29 { 30 long cc=0; 31 vis[d]=true; 32 struct node *p=point[d]; 33 while (p) 34 { 35 if (!vis[p->d]) 36 { 37 cc++; 38 dfs(p->d); 39 need[d]=need[d] ^ need[p->d]; 40 } 41 else 42 fa[d]=p->d; 43 p=p->next; 44 } 45 } 46 47 void print(long d) 48 { 49 printf("%ld\n",d); 50 struct node *p=point[d]; 51 while (p) 52 { 53 if (fa[d]!=p->d && need[p->d]==1) 54 print(p->d); 55 p=p->next; 56 } 57 } 58 59 void work(long d) 60 { 61 struct node *p=point[d]; 62 struct node *q; 63 while (p) 64 { 65 if (fa[d]!=p->d) 66 work(p->d); 67 p=p->next; 68 } 69 if (need[d]==0) 70 { 71 printf("%ld\n",d); 72 p=point[d]; 73 while (p) 74 { 75 if (fa[d]!=p->d && need[p->d]==1) 76 print(p->d); 77 p=p->next; 78 } 79 } 80 } 81 82 int main() 83 { 84 struct node *p; 85 long n,root,i,y; 86 scanf("%ld",&n); 87 88 for (i=1;i<=n;i++) 89 { 90 point[i]=NULL; 91 need[i]=1; //0 92 vis[i]=false; 93 } 94 for (i=1;i<=n;i++) 95 { 96 scanf("%ld",&y); 97 if (y==0) 98 root=i; 99 else 100 { 101 p=(struct node *) malloc (sizeof(struct node)); 102 p->d=y; 103 p->next=point[i]; 104 point[i]=p; 105 106 p=(struct node *) malloc (sizeof(struct node)); 107 p->d=i; 108 p->next=point[y]; 109 point[y]=p; 110 } 111 } 112 need[root]=0; 113 114 fa[root]=0; 115 dfs(root); 116 if (need[root]==1) 117 { 118 printf("NO"); 119 return 0; 120 } 121 122 printf("YES\n"); 123 work(root); 124 125 return 0; 126 } 127 /* 128 9 129 0 1 1 2 2 2 3 6 6 130 131 9 132 0 1 1 1 3 3 3 6 6 133 134 5 135 0 1 2 3 4 136 */