标签：print one dal nsis 4.4 tween clu begin sam
The preferred way to generate user login in Polygon is to concatenate a prefix of the user‘s first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically（字母顺序地） earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix（前缀） of a string s is its substring which occurs at the beginning of s: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and ‘bc" are not. A string a is alphabetically earlier than a string b, if a is a prefix of b, or a and b coincide up to some position, and then a has a letter that is alphabetically earlier than the corresponding letter in b: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
harry potter
hap
tom riddle
tomr
题目意思：给你用户的名和姓，让你生成一个按字母顺序的最短登录名，原则是：名（也就是a串）和姓（也就是b串）都要保留第一个字母，a串除了第一的字母之后的要取到字典序不大于b串第一个字母位置。
说起来还是很繁琐，其实这个有点生活气息，我们平时上网也有这样处理用户名字的，当时我是第二个出这道题的，多少是有点猜的，不过猜对了。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 char a[100],b[100]; 6 char c[100]; 7 int main() 8 { 9 int i,lena,lenb,counts; 10 scanf("%s",&a); 11 scanf("%s",&b); 12 counts=1; 13 lena=strlen(a); 14 lenb=strlen(b); 15 for(i=1;i<lena;i++) 16 { 17 if(a[i]<b[0]) 18 { 19 counts++; 20 } 21 else 22 { 23 break; 24 } 25 } 26 for(i=0;i<counts;i++) 27 { 28 c[i]=a[i]; 29 } 30 c[i]=b[0]; 31 i++; 32 c[i]=‘\n‘; 33 printf("%s",c); 34 return 0; 35 }
Mishka has got n empty boxes. For every i (1 ≤ i ≤ n), i-th box is a cube with side length ai.
Mishka can put a box i into another box j if the following conditions are met:
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
The first line contains one integer n (1 ≤ n ≤ 5000) — the number of boxes Mishka has got.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the side length of i-th box.
Print the minimum possible number of visible boxes.
3
1 2 3
1
4
4 2 4 3
2
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
题目意思：套盒子，尺寸大的盒子能装下尺寸小的盒子，给你一些盒子，问你最后最少能剩下多少的盒子。
解题思路：这个题有点俄罗斯套娃的意思，大的能装下小的，那么我们只需要看看最后剩下尺寸最大的盒子的个数就行了。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 int a[5010]; 6 int main() 7 { 8 int n,counts,maxs,i; 9 scanf("%d",&n); 10 for(i=0; i<n; i++) 11 { 12 scanf("%d",&a[i]); 13 } 14 sort(a,a+n); 15 counts=1; 16 maxs=1; 17 for(i=1; i<n; i++) 18 { 19 if(a[i]==a[i-1]) 20 { 21 counts++; 22 if(maxs<counts) 23 { 24 maxs=counts; 25 } 26 } 27 else 28 { 29 counts=1; 30 } 31 } 32 printf("%d\n",maxs); 33 return 0; 34 }
这里再给出使用map这一容器的做法：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<map> 5 using namespace std; 6 int main() 7 { 8 int n,maxs,i,num; 9 maxs=-1; 10 map<int,int>mp; 11 scanf("%d",&n); 12 for(i=0; i<n; i++) 13 { 14 scanf("%d",&num); 15 mp[num]++; 16 maxs=max(maxs,mp[num]); 17 } 18 printf("%d\n",maxs); 19 return 0; 20 }
标签：print one dal nsis 4.4 tween clu begin sam
原文地址：https://www.cnblogs.com/wkfvawl/p/9409730.html
