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
def q_(x):
a = 8
ans = 1
while(x>0):
if(x&1):
ans = ans*a%10
a = a*a%10
x = x//2
print(ans%10) c =int(input())
q_(c)
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
def gcd(x,y):
return x if y==0 else gcd(y,x%y)
def lcm(x,y):
return x//gcd(x,y)*y
def dfs(x,y):
if(vis[x]):
return y
vis[x] = 1
return dfs(c[x-1],y+1)
n =int(input())
c = []
for x in input().split():
c.append(int(x))
vis = [0]*(n+1)
ind = [0]*(n+1)
for i in range(n):
ind[i+1]+=1
ind[c[i]]+=1
mk = 1
for i in range(n):
if(ind[i+1]&1):
mk = 0
ans = 1
if(mk):
for i in range(n):
if(vis[i+1]==0):
val = dfs(i+1,0)
if(val%2==0):
val//=2
ans = lcm(ans,val)
print(ans)
else:
print("-1");
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[],a[],b[],f[],q[];
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=;i<=n;i++) scanf("%d",&a[i]);
for(i=;i<=n;i++) scanf("%d",&b[i]);
for(i=;i<=m;i++)
{
scanf("%d%d",&r1,&r2);
j=find(r1); k=find(r2);
if(j!=k)fa[k]=j;
}
for(i=;i<=n;i++)
if(fa[i]==)
{
cnt=; int ww=,bb=;
for(j=;j<=n;j++) if(find(j)==i)
{
q[++cnt]=j; ww+=a[j]; bb+=b[j];
}
for(j=w;j>=;j--)
{
//f[i][j]=f[i-1][j];
for(k=;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 ;
}
AC代码
待补
E.
F.