USACO 3.4

洛谷 1827 美国血统

题目

给出一棵树的前序遍历和中序遍历，求它的后序遍历

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代码

/*
ID:lemondi1
LANG:C++
TASK:heritage
*/
#include <cstdio>
#include <cstring>
#define rr register
using namespace std;
char a[31],b[31]; int pos[31];
inline void dfs(int l,int r,int L,int R){
	if (l>r||L>R) return;
	rr int now=pos[a[l]];
	dfs(l+1,l+now-L,L,now-1);
	dfs(l+now-L+1,r,now+1,R);
	putchar(a[l]);
}
signed main(){
	freopen("heritage.in","r",stdin);
	freopen("heritage.out","w",stdout);
	scanf("%s%s",b+1,a+1);
	rr int len=strlen(a+1);
	for (rr int i=1;i<=len;++i) pos[b[i]]=i;
	dfs(1,len,1,len);
	return !putchar(10);
}

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洛谷 2735 电网

题目

有一个由$(0,0),(n,m),(p,0)$组成的三角形，问三角形内部的格点数

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分析

根据皮克定理，可以得到$s=a+\frac{1}{2}b-1$，$a$表示内部的格点，$b$表示边上的格点，$s$表示多边形的面积，那么$a=s-\frac{1}{2}b+1$,所以就可以$O(1)$求

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代码

/*
ID:lemondi1
LANG:C++
TASK:fence9
*/
#include <cstdio>
using namespace std;
inline signed aabs(int x){return x<0?-x:x;}
inline signed gcd(int x,int y){return y?gcd(y,x%y):x;}
signed main(){
	freopen("fence9.in","r",stdin);
	freopen("fence9.out","w",stdout);
	int n,m,p; scanf("%d%d%d",&n,&m,&p);
	return !printf("%d\n",((p*(m-1)-gcd(n,m)-gcd(aabs(n-p),m))>>1)+1);
}

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洛谷 2736 破锣摇滚乐队

题目

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代码

/*
ID:lemondi1
LANG:C++
TASK:rockers
*/
#include <cstdio>
#include <cctype>
#define rr register
using namespace std;
int n,lim,m,ans,a[21];
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline signed max(int a,int b){return a>b?a:b;}
inline void dfs(int dep,int now,int tot,int sum){
	if (dep>n||tot>m){
		ans=max(ans,sum);
		return;
	}
	if (lim>=now+a[dep]){
		if (lim>now+a[dep]) dfs(dep+1,now+a[dep],tot,sum+1);
		dfs(dep+1,0,tot+1,sum+1);//新一个碟子
	}
	dfs(dep+1,now,tot,sum);
}
signed main(){
	freopen("rockers.in","r",stdin);
	freopen("rockers.out","w",stdout);
    n=iut(),lim=iut(),m=iut();
    for (rr int i=1;i<=n;++i) a[i]=iut();
    dfs(1,0,1,0);
    return !printf("%d\n",ans);
}