CF-1082（渣渣只做了前三个）

链接：http://codeforces.com/contest/1082

A. Vasya and Book

题意：

n,x,y,d
一本电子书有n页，每一次翻动只能往前或者往后翻d页。求x->y页最少需要多少步。只能在（1~n)之间翻。具体细节看题目吧。博客仅作记录

int t,n,x,y,d;
int calc(int x,int y)
{
	return abs(x-y)/d;
}
int main() 
{
    scanf("%d",&t);
    while(t--)
    {
    	scanf("%d%d%d%d",&n,&x,&y,&d);
    	int res = abs(x-y);
    	if(res%d==0)
    	{
    		cout<<res/d<<endl;continue;
    	}
    	int ans = inf;
    	if((y-1)%d==0)
    	{
    		ans = min(ans,(int)ceil((x-1.)/d)+(y-1)/d);
    	}
    	if((n-y)%d==0)
    	{
    		ans = min(ans,(int)ceil((double)(n-x)/d)+(n-y)/d);
    	}
    	if(ans == inf)
    		ans = -1;
    	cout<<ans<<endl;
    }
    return 0;
}

B. Vova and Trophies

渣渣的代码：

char s[200010];
int d[200010];
int n;
int main() 
{
    scanf("%d",&n);
    scanf("%s",s+1);
    int num = 0;
    for(int i=1;i<=n;i++)
    	if(s[i]=='G')
    		num++;
    int ans = 0;
    for(int i=1;i<=n;i++)
    {
    	if(s[i]!='G')
    	{
    		d[i] = 0;
    	}
    	else
    	{
    		d[i] = 1;
    		if(s[i-1] == 'G')
    			d[i] = d[i-1]+1;
    	}
    }
    for(int i=1;i<=n;i++)
    {
    	ans = max(ans,d[i]);
    	if(s[i-d[i]]=='S'&&d[i]<num)
    	{
    		if(s[i-d[i]-1]=='G')
    		{
    			if(d[i]+d[i-d[i]-1]<num)
    			ans = max(ans,d[i]+1+d[i-d[i]-1]);
    			else
    				ans = max(ans,d[i]+d[i-d[i]-1]);
    		}
    		ans = max(ans,d[i]+1);
    	}
    }
    printf("%d\n",ans);
    return 0;
}

大神的代码

#include <bits/stdc++.h>
using namespace std;
int res,pre,cnt,g;
int main()
{
	int n;
	cin>>n;
	while(n--)
	{
		char s;
		cin>>s;
		if(s=='G')cnt++,g++;
		else pre = cnt,cnt = 0;
		res = max(res,cnt+pre+1);
	}
	cout<<min(g,res);
	return 0;
}

C. Multi-Subject Competition

vector<int> s[100010];
int n,m;

bool cmp(int a,int b)
{
	return a>b;
}
bool cmp2(vector<int>a,vector<int> b)
{
	return a.size()>b.size();
}
int main() 
{
    cin>>n>>m;
    int t,r;
    for(int i=1;i<=n;i++)
    {
    	scanf("%d%d",&t,&r);
    	s[t].push_back(r);
    }
    int mi = 0;
    for(int i=1;i<=m;i++)
    {
    	sort(s[i].begin(),s[i].end(),cmp);
    	for(int j=1;j<s[i].size();j++)
    		s[i][j] += s[i][j-1];
    	mi = max(mi,(int)s[i].size());
    }
    sort(s+1,s+m+1,cmp2);
    int ans = 0;
    for(int i=0;i<mi;i++)
    {
    	int sum = 0;
    	for(int j=1;j<=m;j++)
    	{
    		if(i>=s[j].size())break;
    		sum = max(sum,sum+s[j][i]);
    		//printf("%d %d\n",j,s[j][i]);
    	}
    	//cout<<sum<<endl;
    	ans = max(ans,sum);
    }
    cout<<ans<<endl;
    return 0;
}

大神的代码

#include<bits/stdc++.h>
using namespace std;
long long n,m,s,r,k,c,mx,a[200000],l;
pair<long long,long long> p[200000];
int main()
{
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		cin>>s>>r;
		p[i]={s,-r};
	}
	sort(p,p+n);
	for(int i=0;i<n;i++)
	{
		if(p[i].first!=l)
		{
			k=0; c=0; l=p[i].first;
		}
		c-=p[i].second;
		k++;
		if(c>0)
			a[k]+=c;
		mx=max(mx,a[k]);
	}
	cout<<mx;
	return 0;
}