8.24考试总结

前言

今天考试还是只有Rank5我很畏惧,感觉要被刷了...(绝望,对不起KingBenQi).其实这一套题目我硬盘里都有,只是今天忘记翻了..绝望.

T1

小奇挖矿
这道题目直接贪心就好了(可能是DP吧..)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define ll long long
#define re register
using namespace std;
inline int gi(){
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
double k,c,w,cost[100010];int n,type[100010];
double max(double a,double b){
	return a>b?a:b;
}
int main(){
#ifndef ONLINE_JUDGE
	freopen("explo.in","r",stdin);
	freopen("explo.out","w",stdout);
#endif
	scanf("%d%lf%lf%lf",&n,&k,&c,&w);
	double ans=0;
	for(re int i=1;i<=n;i++){
		type[i]=gi();scanf("%lf",&cost[i]);
	}
	for(re int i=n;i>=1;i--){
		if(type[i]==1)ans=max(ans,ans*(1-0.01*k)+cost[i]*w);
		else ans=max(ans,ans*(1+0.01*c)-cost[i]*w);
	}
	printf("%.2lf
",ans);
	return 0;
}

T2

小奇的数列
由于nzr运用一些玄学的暴力优化使得这道题目不需要平衡树就可以通过.

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<iostream>
#define ll long long
#define re register
using namespace std;
inline ll gi(){
	ll f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
ll n,q,a[500010],sum[500010];
bool bj[510];
int main(){
#ifndef ONLINE_JUDGE
	freopen("seq.in","r",stdin);
	freopen("seq.out","w",stdout);
#endif
	n=gi(),q=gi();
	for(re ll i=1;i<=n;i++){
		a[i]=gi();
		sum[i]=sum[i-1]+a[i];
	}
	while(q--){
		ll l=gi(),r=gi(),p=gi();ll ans=1e18;
		if(r-l+1>p){puts("0");continue;}
		memset(bj,0,sizeof(bj));
		ll size=sqrt(p)/2;
		for(re int qujian=l-1;qujian<=r;qujian++){
			for(re ll k=size;k>=0;k--)
				if(bj[(p+sum[qujian]-k)%p]){
					ans=min(ans,k);if(!ans)break;
				}
			if(!ans)break;
			bj[sum[qujian]%p]=1;
		}
		if(ans!=1e18){
			printf("%lld
",ans);
			continue;
		}
		int flag=0;
		for(re ll i=l;i<=r;i++){
			for(re ll j=i;j<=r;j++){
				ans=min(ans,(sum[j]-sum[i-1])%p);
				if(sum[i-1]==sum[j]){
					flag=1;break;
				}
			}
			if(flag)break;
		}
		printf("%lld
",ans);
	}
	return 0;
}

T3

小奇回地球
直接SPFA判负环然后二分答案就出来了...

#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<iostream>
#define ll long long
#define re register
using namespace std;
inline int gi(){
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int Inf=1000000000+10;
struct node{
	int to,nxt,w;
}e[1000010];
int front[110],cnt,mark[110],can[110],dis[110],n;
void Add(int u,int v,int w){
	e[++cnt]=(node){v,front[u],w};front[u]=cnt;
}
void dfscan(int u){
	mark[u]=1;
	for(re int i=front[u];i;i=e[i].nxt)
		if(!mark[e[i].to])dfscan(e[i].to);
}
bool dfs(int u,int mid){
	mark[u]=1;
	for(re int i=front[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(can[v] && dis[v]>dis[u]+e[i].w+mid){
			if(mark[v])return 1;
			dis[v]=dis[u]+e[i].w+mid;
			if(dfs(v,mid))return 1;
		}
	}
	mark[u]=0;
	return 0;
}
int vis[100010];
void spfa(int mid){
	queue<int >Q;
	while(!Q.empty())Q.pop();
	Q.push(1);dis[1]=0;vis[1]=0;
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(re int i=front[u];i;i=e[i].nxt){
			int v=e[i].to;
			if(dis[v]>dis[u]+e[i].w+mid && can[v]){
				dis[v]=dis[u]+e[i].w+mid;
				if(!vis[v]){
					vis[v]=1;Q.push(v);
				}
			}
		}
		vis[u]=0;
	}
}
bool check(int mid){
	for(re int i=1;i<=n;i++)
		if(can[i]){
			for(re int j=1;j<=n;j++)dis[j]=Inf;
			memset(mark,0,sizeof(mark));
			if(dfs(i,mid))return 0;
        }
	for(re int i=1;i<=n;i++){
		dis[i]=Inf;vis[i]=0;
	}
	spfa(mid);
	if(dis[n]<0 || dis[n]==Inf)return 0;
	return 1;
}
int m;
int main(){
	freopen("earth.in","r",stdin);
	freopen("earth.out","w",stdout);
	int T=gi();
	while(T--){
		memset(front,0,sizeof(front));cnt=0;
		n=gi();m=gi();
		for(re int i=1;i<=m;i++){
			int u=gi(),v=gi(),w=gi();
			Add(u,v,w);
		}
		memset(mark,0,sizeof(mark));
		dfscan(1);memset(can,1,sizeof(can));
		for(re int i=1;i<=n;i++)if(!mark[i])can[i]=0;
		for(re int i=1;i<=n;i++)
			if(can[i]){
				memset(mark,0,sizeof(mark));
				dfscan(i);
				if(!mark[n])can[i]=0;
			}
		int l=-100000,r=100000,ans=-1;
		while(l<=r){
			int mid=(l+r)/2;
			if(check(mid)){
				ans=dis[n];r=mid-1;
			}
			else l=mid+1;
		}
		printf("%d
",ans);
	}
	return 0;
}