Codeforces Round #383 _python作死系列

A. Arpa’s hard exam and Mehrdad’s naive cheat

题意求1378的n次方的最后一位，懒的写循环节 瞎快速幂

py3 int和LL 合并为int了

 1 def q_(x):
 2     a = 8
 3     ans = 1
 4     while(x>0):
 5         if(x&1):
 6             ans = ans*a%10
 7         a = a*a%10
 8         x = x//2
 9     print(ans%10)
10 
11 c =int(input())
12 q_(c)
View Code

 

B. Arpa’s obvious problem and Mehrdad’s terrible solution

ZZ题  读题浪费了很长时间

C. Arpa's loud Owf and Mehrdad's evil plan

找环  不合法的情况肯定是度为单数的点 然后 DFS就ok  偶数/2 求lcm

 1 def gcd(x,y):
 2     return x if y==0 else gcd(y,x%y)
 3 def lcm(x,y):
 4     return x//gcd(x,y)*y
 5 def dfs(x,y):
 6     if(vis[x]):
 7         return y
 8     vis[x] = 1
 9     return dfs(c[x-1],y+1)
10 n =int(input())
11 c = []
12 for x in input().split():
13     c.append(int(x))
14 vis = [0]*(n+1)
15 ind = [0]*(n+1)
16 for i in range(n):
17     ind[i+1]+=1
18     ind[c[i]]+=1
19 mk = 1
20 for i in range(n):
21     if(ind[i+1]&1):
22         mk = 0
23 ans = 1
24 if(mk):
25     for i in range(n):
26         if(vis[i+1]==0):
27             val = dfs(i+1,0)
28         if(val%2==0):
29             val//=2
30         ans = lcm(ans,val)
31     print(ans)
32 else:
33     print("-1");
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D. Arpa's weak amphitheater and Mehrdad's valuable Hoses

约妹纸，要么约集合，要么约集合中的一个（可用list优化） 并查集加01背包

但是问题来了...这个题做了整整五个小时没A...................  TLE到死         python的多重循环比C++慢百倍....怪不得DE这种题一般没python的出现.... 

优化了range  继续TLE  直到现在  我放弃了 

虽然红了两页  但是学到了不少....蛮阿Q的...

n, m, w = map(int, input().split())
W = []
B = []
W = list(map(int, input().split()))
B = list(map(int, input().split()))
par = [0]*(n+10)
for i in range(0, n):
    par[i] = i
def find(x):
    if(x == par[x]):
        return x
    else:
        par[x] = find(par[x])
        return par[x]
def unite(x, y):
    x = find(x)
    y = find(y)
    if(x != y):
        par[x] = y
for i in range(m):
    x, y = map(int, input().split())
    unite(x-1, y-1)
dp = [0]*(w+10)
q = [0]*(n+10)
for i in range(n):
    if(i==find(i)):
        cnt = 0
        V = 0
        E = 0
        for k in range(n):
            if(find(k)==find(i)):
                q[cnt] = k
                cnt += 1
                V += W[k]
                E += B[k]
        for j in range(w,-1,-1):
            if(j>=V):
                dp[j] = max(dp[j],dp[j-V]+E)
            for k in range(cnt):
                if(j>=W[q[k]]):
                    dp[j] = max(dp[j],dp[j-W[q[k]]]+B[q[k]])
print(dp[w])
理论AC代码

后来去CF上拿了份代码  把python翻译成C++   就AC了

#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,w,cnt;
int fa[1100],a[1100],b[1100],f[1100],q[1100];
int find(int x)
{
    if(!fa[x]) return x;
    fa[x]=find(fa[x]);
    return fa[x];
}
int main()
{
    int i,r1,r2,j,k;
    scanf("%d%d%d",&n,&m,&w);
    for(i=1;i<=n;i++) scanf("%d",&a[i]);
    for(i=1;i<=n;i++) scanf("%d",&b[i]);
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&r1,&r2);
        j=find(r1); k=find(r2);
        if(j!=k)fa[k]=j;
    }
    for(i=1;i<=n;i++) 
        if(fa[i]==0)
        {
            cnt=0; int ww=0,bb=0;
            for(j=1;j<=n;j++) if(find(j)==i) 
            {
                q[++cnt]=j; ww+=a[j]; bb+=b[j];
            }
            for(j=w;j>=0;j--) 
            {
                //f[i][j]=f[i-1][j];
                for(k=1;k<=cnt;k++) if(j>=a[q[k]]) f[j]=max(f[j],f[j-a[q[k]]]+b[q[k]]);
                if(j>=ww) f[j]=max(f[j],f[j-ww]+bb);
            }
        } 

    printf("%d",f[w]);
    return 0;
}
AC代码

待补
E.

F.

 

 

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
posted @ 2016-12-08 15:22  zxMrlc  阅读(200)  评论(0)  编辑  收藏  举报