20191010平安科技，20191016美团

平安科技

1.求平方根

二分法：

a,b= input().strip().split()
a=float(a)
b=int(b)
low=0
high=a
ans=(low+high)/2
for i in range(1000):
    if pow(ans,b)==a:
        break
    if pow(ans,b)>a:
        high=ans
    else:
        low=ans
    ans=(low+high)/2
ans=("%.6f"%ans)
print(ans)

3.对比版本号

注意版本号每部分可能有前导0，应该忽略，返回比较结果

v1,v2=map(str, input().strip().split())
def compare_version(version1,version2):
    print(version1,version2)
    if version1 == version2:
        return 0
    try:
        current_section_version1 = version1[:version1.index('.')]
    except:
        current_section_version1 = version1
    try:
        current_section_version2 = version2[:version2.index('.')]
    except:
        current_section_version2 = version2
    if int(current_section_version1) > int(current_section_version2):
        return 1
    elif int(current_section_version1) < int(current_section_version2):
        return -1
    try:
        other_section_version1 = version1[version1.index('.')+1:]
    except:
        other_section_version1 = "0"
    try:
        other_section_version2 = version2[version2.index('.')+1:]
    except:
        other_section_version2 = "0"
    return compare_version(other_section_version1,other_section_version2)
print(compare_version(v1,v2))

美团

1.

根据划分方法，所有的字符串按照首字母排序并进行划分，划分之后对每个部分继续对下一位进行排序 

arr = list(map(str, input().strip().split(',')))
""" arr.sort(reverse=True)
print(','.join(arr)) """
#waimai,dache,lvyou,liren,meishi,jiehun,lvyoujingdian,jiaopei,menpiao,jiudian
#waimai,menpiao,meishi,lvyou,lvyoujingdian,liren,jiudian,jiehun,jiaopei,dache
def charat(s,i):
    n=len(s)
    if i<n:
        return ord('z')-ord(s[i])
    else:
        return -1
def sortnew(arr):
    n=len(arr)
    aux=['' for i in range(n)]
    sortit(arr,aux,0,n-1,0)
def sortit(arr,aux,lo,hi,d):
    if lo>=hi:
        return
    R=26
    count=[0 for i in range(R+2)]
    for i in range(lo,hi+1):
        count[charat(arr[i],d)+2]+=1
    for i in range(0,R+1):
        count[i+1]+=count[i]
    #print(len(count))
    #print(count)
    for i in range(lo,hi+1):
        x=charat(arr[i],d)+1
        #print(count)
        aux[count[x]]=arr[i]
        count[x]+=1
    for i in range(lo,hi+1):
        arr[i]=aux[i-lo]
    for i in range(0,R):
        sortit(arr,aux,lo+count[i],lo+count[i+1]-1,d+1)
sortnew(arr)
print(','.join(arr))

2.

没全A，算法是有待改进。。。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<sstream>
#include<cmath>
#include<climits>
#include<queue>
#include<cstring>
#include<set>
#include <string>
using namespace std;
typedef struct box {
	int cur;
	int limit;
}box;
bool cmp(const box& a,const box& c){
    if(a.limit==c.limit)
    return a.cur>c.cur;
    return a.limit > c.limit;
}
int main(){
	int n;
	cin >> n;
	vector<box> b(n);
	for (int i = 0; i < n; ++i)
		cin >> b[i].cur;
	for (int i = 0; i < n; ++i)
		cin >> b[i].limit;
	sort(b.begin(), b.end(), cmp);
	int i, res = 0, last = n - 1;
	for (i = 0; i < last; ++i) {
		int le = b[i].limit - b[i].cur;
		if (le < b[last].cur) {
			b[last].cur -= le;
			b[i].cur = b[i].limit;
			res += le;
		} else {
			b[i].cur += b[last].cur;
			res += b[last].cur;
			b[last--].cur = 0;
		}
		if (b[i].cur < b[i].limit)
			--i;
	}
	cout <<i<<' '<<res<<endl;
    system("PAUSE");
	return 0;
}
/*4 
3 3 4 3
4 7 6 5
2 6*/