标签：this row several long 所有路径 view 选择 计算 preview
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2245 Accepted Submission(s): 1053
Problem Description
There is a youngster known for amateur propositions concerning several mathematical hard problems.
Nowadays, he is preparing a thought-provoking problem on a specific type of supercomputer which has ability to support calculations of integers between 0
and (2
m
?1)
(inclusive).
As a young man born with ten fingers, he loves the powers of 10
so much, which results in his eccentricity that he always ranges integers he would like to use from 1
to 10
k
(inclusive).
For the sake of processing, all integers he would use possibly in this interesting problem ought to be as computable as this supercomputer could.
Given the positive integer m
, your task is to determine maximum possible integer k
that is suitable for the specific supercomputer.
Input
The input contains multiple test cases. Each test case in one line contains only one positive integer m
, satisfying 1≤m≤$10^5$ .
Output
For each test case, output "Case #x
: y
" in one line (without quotes), where x
indicates the case number starting from 1
and y
denotes the answer of corresponding case.
Sample Input
Sample Output
Case #1: 0 Case #2: 19
1 #include<bits/stdc++.h> 2 #define clr(x) memset(x,0,sizeof(x)) 3 using namespace std; 4 int main() 5 { 6 int m; 7 int t = 1; 8 while(cin >> m) { 9 10 cout << "Case #" << t ++ << ": " << (int)(m*0.30102999566)<< "\n"; 11 } 12 return 0; 13 }
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4809 Accepted Submission(s): 387
Problem Description
Talented Mr.Tang has n
strings consisting of only lower case characters. He wants to charge them with Balala Power (he could change each character ranged from a to z into each number ranged from 0 to 25, but each two different characters should not be changed into the same number) so that he could calculate the sum of these strings as integers in base 26
hilariously.
Mr.Tang wants you to maximize the summation. Notice that no string in this problem could have leading zeros except for string "0". It is guaranteed that at least one character does not appear at the beginning of any string.
The summation may be quite large, so you should output it in modulo 109
+7
.
Input
The input contains multiple test cases.
For each test case, the first line contains one positive integers n
, the number of strings. (1≤n≤100000)
Each of the next n
lines contains a string s
i
consisting of only lower case letters. (1≤|s
i
|≤100000,∑|s
i
|≤10
6
)
Output
For each test case, output "Case #x
: y
" in one line (without quotes), where x
indicates the case number starting from 1
and y
denotes the answer of corresponding case.
Sample Input
Sample Output
Case #1: 25 Case #2: 1323 Case #3: 18221
1 #include <bits/stdc++.h> 2 #define clr(x) memset(x,0,sizeof(x)) 3 #define mod 1000000007 4 #define LL long long 5 using namespace std; 6 char ss[100020]; 7 int n,m,k,l,t,minc,len; 8 int charc[50][100020]; 9 bool firsted[200]; 10 int inf[200]; 11 bool comp(int a,int b) 12 { 13 for(int i=m;i>=0;i--) 14 if(charc[a-‘a‘][i]!=charc[b-‘a‘][i]) 15 return charc[a-‘a‘][i]>charc[b-‘a‘][i]; 16 return 0; 17 } 18 int max(int a,int b) 19 { 20 return a>b?a:b; 21 } 22 int main() 23 { 24 int T,kase=0; 25 LL ans,mul; 26 while(scanf("%d",&n)!=EOF) 27 { 28 m=0; 29 clr(firsted); 30 clr(charc); 31 printf("Case #%d: ",++kase); 32 for(int i=1;i<=n;i++) 33 { 34 scanf("%s",ss); 35 len=strlen(ss); 36 if(len>0) firsted[ss[0]]=1; 37 reverse(ss,ss+len); 38 for(int j=0;j<len;j++) 39 charc[ss[j]-‘a‘][j]++; 40 m=max(m,len); 41 } 42 for(int i=‘a‘;i<=‘z‘;i++) 43 { 44 // cout<<i<<":"; 45 for(int j=0;j<m;j++) 46 { 47 if(charc[i-‘a‘][j]>25) 48 { 49 charc[i-‘a‘][j+1]+=charc[i-‘a‘][j]/26; 50 charc[i-‘a‘][j]%=26; 51 } 52 // cout<<" "<<charc[i-‘a‘][j]; 53 } 54 while(charc[i-‘a‘][m]>25) 55 { 56 charc[i-‘a‘][m+1]+=charc[i-‘a‘][m]/26; 57 charc[i-‘a‘][m]%=26; 58 m++; 59 // cout<<" "<<charc[i-‘a‘][m]; 60 } 61 if(charc[i-‘a‘][m]>0) 62 m++; 63 // cout<<endl; 64 } 65 clr(inf); 66 ans=0; 67 for(int i=0;i<=25;i++) 68 { 69 for(int j=‘a‘;j<=‘z‘;j++) 70 if((i==0 && firsted[j]==0) || (i!=0 && inf[j-‘a‘]==0)) 71 { 72 minc=j; 73 break; 74 } 75 for(int j=‘a‘;j<=‘z‘;j++) 76 if((i==0 && firsted[j]==0) || (i!=0 && inf[j-‘a‘]==0)) 77 if(!comp(j,minc)) 78 minc=j; 79 mul=1; 80 for(int j=0;j<=m;j++) 81 { 82 ans=(ans+(LL)charc[minc-‘a‘][j]*mul*(LL)i)%mod; 83 mul=(mul*26)%mod; 84 } 85 inf[minc-‘a‘]=1; 86 // cout<<ans<<" "<<maxc<<endl; 87 } 88 printf("%lld\n",ans); 89 } 90 return 0; 91 }
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 392 Accepted Submission(s): 68
Problem Description
There is a tree with n
nodes, each of which has a type of color represented by an integer, where the color of node i
is c
i
.
The path between each two different nodes is unique, of which we define the value as the number of different colors appearing in it.
Calculate the sum of values of all paths on the tree that has n(n?1)2
paths in total.
Input
The input contains multiple test cases.
For each test case, the first line contains one positive integers n
, indicating the number of node. (2≤n≤200000)
Next line contains n
integers where the i
-th integer represents c
i
, the color of node i
. (1≤c
i
≤n)
Each of the next n?1
lines contains two positive integers x,y
(1≤x,y≤n,x≠y)
, meaning an edge between node x
and node y
.
It is guaranteed that these edges form a tree.
Output
For each test case, output "Case #x
: y
" in one line (without quotes), where x
indicates the case number starting from 1
and y
denotes the answer of corresponding case.
Sample Input
Sample Output
Case #1: 6 Case #2: 29
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<stack> 5 #include<vector> 6 #define clr(x) memset(x,0,sizeof(x)) 7 #define clr_1(x) memset(x,-1,sizeof(x)) 8 #define LL long long 9 using namespace std; 10 int head[200005],cnt,ncnt,dif,col[200005]; 11 LL ans; 12 int n,m,k,t,s,l; 13 struct edg 14 { 15 int next,to; 16 }edge[400005]; 17 void addedge(int u,int v) 18 { 19 edge[++cnt].next=head[u]; 20 edge[cnt].to=v; 21 head[u]=cnt; 22 return ; 23 } 24 int some[200005]; 25 stack<int> sta[200005]; 26 void dfs(int u,int pred) 27 { 28 some[u]=1; 29 sta[col[u]].push(u); 30 for(int i=head[u];i!=-1;i=edge[i].next) 31 { 32 if(edge[i].to!=pred) 33 { 34 dfs(edge[i].to,u); 35 some[u]+=some[edge[i].to]; 36 dif=some[edge[i].to]; 37 while(sta[col[u]].top()!=u) 38 { 39 dif-=some[sta[col[u]].top()]; 40 sta[col[u]].pop(); 41 } 42 ans+=(LL)dif*(dif-1)/2; 43 } 44 } 45 return ; 46 } 47 int main() 48 { 49 int kase=0,u,v,diff,p; 50 while(scanf("%d",&n)!=EOF) 51 { 52 for(int i=1;i<=n;i++) 53 while(!sta[i].empty()) 54 sta[i].pop(); 55 printf("Case #%d: ",++kase); 56 clr_1(head); 57 cnt=ncnt=0; 58 for(int i=1;i<=n;i++) 59 scanf("%d",&col[i]); 60 for(int i=1;i<n;i++) 61 { 62 scanf("%d%d",&u,&v); 63 addedge(u,v); 64 addedge(v,u); 65 } 66 ans=0; 67 clr(some); 68 dfs(1,1); 69 for(int i=1;i<=n;i++) 70 { 71 diff=n; 72 while(!sta[i].empty()) 73 { 74 p=sta[i].top(); 75 diff-=some[p]; 76 sta[i].pop(); 77 } 78 ans+=(LL)diff*(LL)(diff-1)/2; 79 } 80 ans=(LL)n*n*(n-1)/2-ans; 81 printf("%lld\n",ans); 82 } 83 return 0; 84 }
1 #include<bits/stdc++.h> 2 #define clr(x) memset(x,0,sizeof(x)) 3 #define LL long long 4 using namespace std; 5 int main() 6 { 7 LL n,k,kase=0,t,l,ans; 8 while(scanf("%lld%lld",&n,&k)!=EOF) 9 { 10 printf("Case #%lld: ",++kase); 11 if(k<=n) 12 printf("%lld\n",k); 13 else 14 { 15 k-=n; 16 t=k/(n-1); 17 l=k%(n-1); 18 if(l!=0) 19 printf("%lld\n",l); 20 else if(t&1) 21 printf("%lld\n",n-1); 22 else 23 printf("%lld\n",n); 24 } 25 } 26 return 0; 27 }
2017 Multi-University Training 1 解题报告
标签：this row several long 所有路径 view 选择 计算 preview
原文地址：http://www.cnblogs.com/wujiechao/p/7240164.html
