标签：imu end pretty diff nsis upper == inpu 组成
A. Search for Pretty Integers
You are given two lists of non-zero digits.
Let‘s call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?
The first line contains two integers n and m (1?≤?n,?m?≤?9) — the lengths of the first and the second lists, respectively.
The second line contains n distinct digits a1,?a2,?...,?an (1?≤?ai?≤?9) — the elements of the first list.
The third line contains m distinct digits b1,?b2,?...,?bm (1?≤?bi?≤?9) — the elements of the second list.
Print the smallest pretty integer.
2 3
4 2
5 7 6
25
8 8
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1
1
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don‘t have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It‘s obvious that the smallest among them is 1, because it‘s the smallest positive integer.
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int INF=10; 8 int n,m; 9 10 int main() 11 { int a[10],b[10],visa[10],visb[10]; 12 cin>>n>>m; 13 memset(visa,0,sizeof(visa)); 14 memset(visb,0,sizeof(visb)); 15 for(int i=1;i<=n;i++){ cin>>a[i]; visa[a[i]]=1; } 16 for(int i=1;i<=m;i++){ cin>>b[i]; visb[b[i]]=1; } 17 sort(a+1,a+n+1); 18 sort(b+1,b+m+1); 19 int ans=INF; 20 for(int i=1;i<=9;i++) if(visa[i]&&visb[i]){ ans=i; break; } 21 if(ans!=INF) cout<<ans<<endl; 22 else{ 23 if(a[1]<b[1]) printf("%d%d\n",a[1],b[1]); 24 else printf("%d%d\n",b[1],a[1]); 25 } 26 }
B. Maximum of Maximums of Minimums
You are given an array a1,?a2,?...,?an consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You‘ll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?
Definitions of subsegment and array splitting are given in notes.
The first line contains two integers n and k (1?≤?k?≤?n?≤??105) — the size of the array a and the number of subsegments you have to split the array to.
The second line contains n integers a1,??a2,??...,??an (?-?109??≤??ai?≤??109).
Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.
5 2
1 2 3 4 5
5
5 1
-4 -5 -3 -2 -1
-5
A subsegment [l,??r] (l?≤?r) of array a is the sequence al,??al?+?1,??...,??ar.
Splitting of array a of n elements into k subsegments [l1,?r1], [l2,?r2], ..., [lk,?rk] (l1?=?1, rk?=?n, li?=?ri?-?1?+?1 for all i?>?1) is k sequences (al1,?...,?ar1),?...,?(alk,?...,?ark).
In the first example you should split the array into subsegments [1,?4] and [5,?5] that results in sequences (1,?2,?3,?4) and (5). The minimums are min(1,?2,?3,?4)?=?1 and min(5)?=?5. The resulting maximum is max(1,?5)?=?5. It is obvious that you can‘t reach greater result.
In the second example the only option you have is to split the array into one subsegment [1,?5], that results in one sequence (?-?4,??-?5,??-?3,??-?2,??-?1). The only minimum is min(?-?4,??-?5,??-?3,??-?2,??-?1)?=??-?5. The resulting maximum is ?-?5.
题解：分类讨论，k>=3的情况，只要把数组中最大值单独作为一个区间，那么就得到了ans.
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int INF=2e9; 8 const int maxn=2e5; 9 10 int n,k; 11 int a[maxn]; 12 13 int main() 14 { cin>>n>>k; 15 int mi=INF,ma=-INF; //因为数据可能是负的！ 16 for(int i=1;i<=n;i++){ 17 scanf("%d",&a[i]); 18 mi=min(mi,a[i]); 19 ma=max(ma,a[i]); 20 } 21 if(k==1) cout<<mi<<endl; 22 else if(k==2){ 23 int b[maxn],c[maxn],x=INF; 24 for(int i=1;i<=n;i++){ 25 x=min(x,a[i]); 26 b[i]=x; 27 } 28 x=INF; 29 for(int i=n;i>=1;i--){ 30 x=min(x,a[i]); 31 c[i]=x; 32 } 33 int ans=-INF; 34 for(int i=1;i<n;i++) ans=max(ans,max(b[i],c[i+1])); 35 cout<<ans<<endl; 36 } 37 else cout<<ma<<endl; 38 }
C. Maximum splitting
You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
The first line contains single integer q (1?≤?q?≤?105) — the number of queries.
q lines follow. The (i?+?1)-th line contains single integer ni (1?≤?ni?≤?109) — the i-th query.
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
1
12
3
2
6
8
1
2
3
1
2
3
-1
-1
-1
12?=?4?+?4?+?4?=?4?+?8?=?6?+?6?=?12, but the first splitting has the maximum possible number of summands.
8?=?4?+?4, 6 can‘t be split into several composite summands.
1,?2,?3 are less than any composite number, so they do not have valid splittings
题意：问一个数最多能用几个合数表示
题解：如果是偶数，那么它能表示成 n=k*2(k个2相乘),答案就是k/2,即只由4和6组成。如果是奇数，那么它能表示成 n=k*2+1，合数由4，6，9组成，所以要分类讨论
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int kase,n; 5 6 int main() 7 { scanf("%d",&kase); 8 while(kase--){ 9 scanf("%d",&n); 10 if(n==1||n==2||n==3){ cout<<"-1"<<endl; continue; } 11 if(n%2==0) cout<<n/2/2<<endl; 12 else{ 13 int ans=n/2-4; 14 if(ans==0) cout<<"1"<<endl; 15 else if(ans<0||ans==1) cout<<"-1"<<endl; 16 else cout<<ans/2+1<<endl; //别忘了加1(即合数9)!!! 17 } 18 } 19 return 0; 20 }
标签：imu end pretty diff nsis upper == inpu 组成
原文地址：http://www.cnblogs.com/zgglj-com/p/7674413.html
