T1

题解：这道题属于水题，模拟

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
long long int n,ans1=-10,ans2,a,b,c,an,na,ns;
char s[21],as[21],st1,st2;
void getc()
{
	na=1;
	scanf("%c",&s[na]);
	while(s[na]!=' ')scanf("%c",&s[++na]);
	na--;
}
int main()
{
	freopen("scholar.in","r",stdin);
	freopen("scholar.out","w",stdout);
	scanf("%lld\n",&n);
	for(int i=1;i<=n;i++)
	{
		getc();
		scanf("%lld %lld %c %c %lld\n",&a,&b,&st1,&st2,&c);
		an=0;
		if(a>80&&c>=1)an+=8000;
		if(a>85&&b>80)an+=4000;
		if(a>90)an+=2000;
		if(a>85&&st2=='Y')an+=1000;
		if(b>80&&st1=='Y')an+=850;
		ans2+=an;
		if(an>ans1)
		{
			ans1=an;
			ns=na;
			for(int j=1;j<=na;j++)
			{
				as[j]=s[j];
			}
		}
	}
	for(int i=1;i<=ns;i++)
	{
		 printf("%c",as[i]);
	}printf("\n");
	printf("%lld\n%lld",ans1,ans2);
	return 0;
}

T2

题解：离散化压位，搞一下就行了

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
long long int l,s,t,m,f[1000001],st[150000],fl[1500000],ans=999999999;
int main()
{
	freopen("river.in","r",stdin);
	freopen("river.out","w",stdout);
	memset(f,127,sizeof(f));
	scanf("%lld%lld%lld%lld",&l,&s,&t,&m);
	for(long long int i=1;i<=m;i++)scanf("%lld",&st[i]);
	sort(st+1,st+m+1);
	if(s==t)
	{
		ans=0;
		for(int i=1;i<=m;i++)if(st[i]%s==0)ans++;
		printf("%lld",ans);return 0;
	}
	st[0]=0;
	long long int k=s*t/(t-s)+s+t+1;l=0;
	for(long long int i=1;i<=m;i++)
	{
		l+=min(st[i]-st[i-1],k);
		fl[l]=1;
	}
	f[0]=0;
	for(int i=0;i<=l+t;i++)
	{
		for(int j=i-t;j<=i-s;j++)
		{
			if(j>=0)
			{
				f[i]=min(f[i],f[j]+fl[i]);
			}
		}
	}
	for(int i=l;i<=l+t;i++)
	{
		ans=min(ans,f[i]);
	}
	printf("%lld",ans);
	return 0;
}

T3

题解：建个目标环，顺时针和逆时针搜一遍当前点与目标的距离，然后找距离一致中的最多的，用n减去它就是答案

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
long long int n,a[50001],b[50001],dis[100001],dis2[100001],ans=-2147483647;
struct nod{long long int l,r;}f[50001];
int main()
{
	freopen("fire.in","r",stdin);
	freopen("fire.out","w",stdout);
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
	{
		a[i]=i;
		scanf("%lld %lld",&f[i].l,&f[i].r);
	}
	b[1]=1;
	b[2]=f[1].l;
	b[n]=f[1].r;
	for(int i=3;i<=n-1;i++)
    {
        if(b[i-2]==f[b[i-1]].l)
        {
            b[i]=f[b[i-1]].r;
        }
        else if(b[i-2]==f[b[i-1]].r)
        {
            b[i]=f[b[i-1]].l;
        }
        else
        {
            printf("-1");
            return 0;
        }
    }
	for(int i=1;i<=n;i++)
	{
		dis[(i-b[i]+n)%n]++;
		dis2[(b[i]+i+n)%n]++;
	}
	for(int i=0;i<=2*n;i++)
	{
		ans=max(ans,max(dis[i],dis2[i]));
	}
	long long int xc=n-ans;
	printf("%lld\n",xc);
	return 0;
}

T4

作者正在写代码中…