20190908爱奇艺，BIGO，字节跳动
爱奇艺
1.
Leetcode原题
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define mod 1000000007
int n;
int dp[1001][1001];
bool flag[1001];
int main(){
scanf("%d",&n);
int s;
for(int i=0;i=0;j--){
dp[i+1][j]=cur=(cur+dp[i][j+1])%mod;
}
}
}
/*for(int i=0;i 2.
红球n个，蓝球m个，ABC三个人，从A开始依次拿球，AB拿到红球游戏结束，C是捣乱的，可以拿红球和蓝球，但是他拿到红球不算赢，如果他把红球拿光了，算B赢，计算A获胜的概率。
超时了。。。
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int t;
int n,m;
double dfs(int abc,int r,int b){
if(r<=0)
return 0.0;
if(abc==0){
if(b==0)
return (double)r/(r+b);
return (double)r/(r+b)+(double)b/(r+b)*dfs(1,r,b-1);
}
else{
if(abc==1){
if(b==0)
return 0.0;
return (double)b/(r+b)*dfs(2,r,b-1);
}
else{
if(b==0)
return (double)r/(r+b)*dfs(0,r-1,b);
return (double)r/(r+b)*dfs(0,r-1,b)+(double)b/(r+b)*dfs(0,r,b-1);
}
}
}
int main(){
scanf("%d %d",&n,&m);
//int sum=n+m;
double ans=1.0;
ans=dfs(0,n,m);
printf("%.5lf\n",ans);
//system("PAUSE");
return 0;
} 可能要用数组暂存优化
#include
#include
#include
#include
#include BIGO
1.
拿石子问题（博弈论）
2.
aligned_malloc
3.
爬n阶楼梯，一次可以爬1,2,3阶，求方法的总数
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int n;
int dfs(int left){
if(left==0)return 1;
if(left<0) return 0;
return dfs(left-1)+dfs(left-2)+dfs(left-3);
}
int climb_base(int n){
int ans=dfs(n);
return ans;
}
int main(){
scanf("%d",&n);
int ans;
ans=climb_base(n);
cout< 4.
求两个不相交的连续子数组的最大和
头条
1.
n = int(input().strip())
a = list(map(int, input().strip().split()))
res = [-1, 0]
i = n-1
stack = [[a[i], i]]
i -= 1
while stack and i >= 0:
ans = 0
while stack and a[i] >= stack[-1][0]:
stack.pop(-1)
ans += 1
if ans >= res[1]:
res = [a[i], ans]
stack.append([a[i], i])
i -= 1
print(res[0])2.倒水
import java.util.*;
public class Main {
public static class Water{
public int l1=0;
public int l2=0;
public int l3=0;
public Water(Water w){
l1=w.l1;l2=w.l2;l3=w.l3;
}
public Water(){
l1=0;l2=0;l3=0;
}
@Override
public int hashCode(){
String str=String.valueOf(l1)+String.valueOf(l2)+String.valueOf(l3);
return Integer.valueOf(str);
}
@Override
public boolean equals(Object obj){
if(obj instanceof Water) {
Water w = (Water) obj;
return (w.l1 == l1 && w.l2 == l2 && w.l3 == l3);
}
return false;
}
}
public static void bfs(int kettle[],Water water,HashSet set,List queue){
Water w=new Water(water);
w.l1=kettle[0];
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l2=kettle[1];
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l3=kettle[2];
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l1=0;
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l2=0;
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l3=0;
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l1=(water.l1+water.l2-kettle[1])<0?0:(water.l1+water.l2-kettle[1]);
w.l2=(water.l1+water.l2>kettle[1])?kettle[1]:(water.l1+water.l2);
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l2=(water.l1+water.l2-kettle[0])<0?0:(water.l1+water.l2-kettle[0]);
w.l1=(water.l1+water.l2>kettle[0])?kettle[0]:(water.l1+water.l2);
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l1=(water.l1+water.l3-kettle[2])<0?0:(water.l1+water.l3-kettle[2]);
w.l3=(water.l1+water.l3>kettle[2])?kettle[2]:(water.l1+water.l3);
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l3=(water.l1+water.l3-kettle[0])<0?0:(water.l1+water.l3-kettle[0]);
w.l1=(water.l1+water.l3>kettle[0])?kettle[0]:(water.l1+water.l3);
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l2=(water.l2+water.l3-kettle[2])<0?0:(water.l2+water.l3-kettle[2]);
w.l3=(water.l2+water.l3>kettle[2])?kettle[2]:(water.l2+water.l3);
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
w=new Water(water);
w.l3=(water.l2+water.l3-kettle[1])<0?0:(water.l2+water.l3-kettle[1]);
w.l2=(water.l2+water.l3>kettle[1])?kettle[1]:(water.l2+water.l3);
if(!set.contains(w)){
set.add(w);
queue.add(w);
}
}
public static int func(int kettle[],int target){
Water water=new Water();
HashSet set=new HashSet<>();
List queue=new ArrayList<>();
queue.add(water);
set.add(water);
int deep=0;
while(!queue.isEmpty()){
int size=queue.size();
for(int i=0;i 3.
4.
ch='0ABCDEFGHIJKLMNOPQRSTUVWXYZ'
s = str(input().strip())
n = len(s)
res=''
def dfs(i,l,res):
if i==l :
print(res)
if l-i>=1:
a=int(s[i])
if a==0:
return
dfs(i+1,l,res+ch[a])
if l-i >= 2 :
a=int(s[i:i+2])
if a<10 or a>=27:
return
dfs(i+2,l,res+ch[a])
dfs(0,n,res)