Codeforces Round #299 (Div. 2)【A,B,C】

codeforces 535A-水题；

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

char s2[15][20]={"eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};

char s1[15][20]={"ten","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"};

char s3[15][20]={"one","two","three","four","five","six","seven","eight","nine"};

int main()

{

	int n;

	scanf("%d",&n);

	if(n==0)

	{

		puts("zero");

		return 0;

	}

	if(n>=10)

	{

		if(n%10==0)

			printf("%s",s1[n/10-1]);

		else if((n/10)==1)

			printf("%s",s2[n%10-1]);

		else

			printf("%s-%s",s1[n/10-1],s3[n%10-1]);

	}

	else

		printf("%s",s3[n%10-1]);

	return 0;

}

codeforces 535B

最多才2^9个数，直接预处理，然后找一遍。

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

LL p[1000000];

LL Pow(int k)

{

	LL ans=1;

	for(int i=1;i<=k;i++)

		ans=ans*10LL;

	return ans;

}

int num;

void init()

{

	LL k;

	p[0]=4LL;

	p[1]=7LL;

	num=2;

	bool flag=false;

	for(int i=1;i<=9;i++)

	{

		int t=0;

		k=Pow(i);

		for(int j=0;j<num;j++)

		{

			if(p[j]*10<k) continue;

			p[num+t]=k*4LL+p[j];

			if(p[num+t]>1000000000)

			{

				num=num+t;

				flag=true;

				break;

			}

			t++;

			p[num+t]=k*7LL+p[j];

			if(p[num+t]>1000000000)

			{

				num+=t;

				flag=true;

				break;

			}

			t++;

		}

		if(flag) break;

		num=num+t;

	}

}

int main()

{

	init();

	LL n;

	scanf("%lld",&n);

	sort(p,p+num);

	for(int i=0;i<num;i++)

	{

		if(p[i]==n)

		{

			printf("%d\n",i+1);

			return 0;

		}

	}

	return 0;

}

codeforces 535C：题意：

给等差数列：首项A，公差B，n个询问

每个询问给l,t,m

以l为左端点，每次可以最多选择m个数，使这些数 -1

t次操作后，求最长序列使所有数为0，输出这个最长序列的右端序号

思路：

二分对吧。

那么就去找二分的满足条件对吧。

首先数列里面最大的数肯定<=t，对勾

数列里面所有的数相加和<=t*m 对勾

然后二分 11111100000型对勾

#include <cstdio>

#include <iostream>

#include <stdlib.h>

#include <string.h>

#include <queue>

#include <algorithm>

using namespace std;

typedef long long LL;

int n;

LL a,b,t,m;

LL sum[1000100];

void init()

{

	sum[0]=0;

	sum[1]=a;

	for(LL i=2;i<=1000000;i=i+1LL)

		sum[i]=sum[i-1]+a+(i-1LL)*b;

}

int main()

{

	LL tmp;

	scanf("%I64d%I64d%I64d",&a,&b,&n);

	init();

	while(n--)

	{

		scanf("%I64d%I64d%I64d",&tmp,&t,&m);

		LL left=tmp,right=1000000;

		while(left<right)

		{

			LL mid=left+(right-left+1LL)/2LL;

			if((a+(mid-1LL)*b)<=t&&(sum[mid]-sum[tmp-1])<=t*m)

				left=mid;

			else

				right=mid-1LL;

		}

		if((a+(left-1LL)*b)<=t&&(sum[left]-sum[tmp-1])<=t*m)

			printf("%I64d\n",left);

		else

			puts("-1");

	}

	return 0;

}

/*

2 1 4

1 5 3

3 3 10

7 10 2

6 4 8

*/

巴特西

Codeforces Round #299 (Div. 2)【A,B,C】

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