Andrew Stankevich Contest 45 (ASC 45)
作者:互联网
经典队内倒二啊 十分经典 感觉有蛮久没训练了 一打起来十分拉跨
Problem A CFGym 100520A Analogous Sets +++
Problem B CFGym 100520B Bayes' Law ???
Problem C CFGym 100520C Catalian Sequences ???
Problem D CFGym 100520D Drunkard's Walk !!!
正解的想法我们想过的 不知道为啥把它hack了 丢
构造一棵满二叉树然后分配叶子结点就行
其中 p个指向home q-p个指向bar 剩余的指向1号结点
这样的话流量最终都只会流向那q个结点
#include<bits/stdc++.h>
using namespace std;
const int N = 300;
int s[15],p2[15];
#define LOCAL
int main(){
#ifdef LOCAL
freopen("drunkard.in", "r", stdin);
freopen("drunkard.out", "w", stdout);
#endif
int p,q;
p2[0] = 1;
s[0] = 1;
for(int i = 1; i <= 10; i++){
p2[i] = p2[i-1]*2;
s[i] = s[i-1] + p2[i];
}
while(~scanf("%d%d",&p,&q)){
if(p==0&&q==0) break;
int j;
for(int i = 1; i <= 10; i++){
if(q <= p2[i]){
j = i;
break;
}
}
int n = s[j] + 2;
printf("%d\n",n);
for(int i = 1; i <= n-2-p2[j]; i++){
printf("%d %d\n",i*2,i*2+1);
}
int now = n-2-p2[j]+1;
for(int i = now; i <= now+p-1; i++){
printf("%d %d\n",n-1,n-1);
}
now+=p;
for(int i = now; i <= now+q-p-1; i++){
printf("%d %d\n",n,n);
}
now+=q-p;
for(int i = now; i <= n-2; i++){
printf("%d %d\n",1,1);
}
}
return 0;
}
Problem E CFGym 100520E Elegant Scheduling ???
Problem F CFGym 100520F Flights !!!
Problem G CFGym 100520G Genome of English Literature ???
Problem H CFGym 100520H Hide-and-Seek ???
Problem I CFGym 100520I Informatics Final Project ???
Problem J CFGym 100520J Japanese Origami ???
Problem K CFGym 100520K Kabbalah for Two !!!
标签:结点,指向,int,45,Stankevich,ASC,freopen,Problem,CFGym 来源: https://www.cnblogs.com/League-of-cryer/p/13985478.html