Codeforces Round #308 (Div. 2)
A. Vanya and Table
Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom to top, the columns are numbered from 1 to 100 from left to right.
In this table, Vanya chose n rectangles with sides that go along borders of squares (some rectangles probably occur multiple times). After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result.
The first line contains integer n (1 ≤ n ≤ 100) — the number of rectangles.
Each of the following n lines contains four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ 100, 1 ≤ y1 ≤ y2 ≤ 100), where x1 and y1 are the number of the column and row of the lower left cell and x2 and y2 are the number of the column and row of the upper right cell of a rectangle.
In a single line print the sum of all values in the cells of the table.
2
1 1 2 3
2 2 3 3
10
2
1 1 3 3
1 1 3 3
18
Note to the first sample test:
Values of the table in the first three rows and columns will be as follows:
121
121
110
So, the sum of values will be equal to 10.
Note to the second sample test:
Values of the table in the first three rows and columns will be as follows:
222
222
222
So, the sum of values will be equal to 18.
题意:给你n个矩形,问你面积之和
转载请注明出处:寻找&星空の孩子
题目链接: http://codeforces.com/contest/552/problem/A
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include<string>
#include<map>
using namespace std;
const int MAXN = ;
int a[][]; int main()
{
int i,j,k,n,m,x1,x2,y1,y2,ans = ;
int maxx = ,maxy=,minx=,miny=;
scanf("%d",&n);
memset(a,,sizeof(a));
while(n--)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
/* maxx = max(maxx,max(x1,x2));
maxy = max(maxy,max(y1,y2));
minx = max(minx,max(x1,x2));
miny = max(miny,max(y1,y2));*/
for(i = x1;i<=x2;i++)
{
for(j = y1;j<=y2;j++)
ans++;
}
}
printf("%d\n",ans);
// int ans = 0; return ;
}
B. Vanya and Books
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.
Print the number of digits needed to number all the books.
13
17
4
4
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.
题意 :给你一个整数n,问你从1到n一共有多少位。比如n = 13,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13一共17位。
f[i]表示从1到10i一共有多少位,以753为例,从100到753都是3位数,所以答案就是f[2]+653*3。
转载请注明出处:寻找&星空の孩子
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include<string>
#include<map>
#define LL __int64
using namespace std;
const int MAXN = ;
LL n;
LL ans = ;
LL a[]= {,,,,,,,,,,}; int main()
{
while(~scanf("%I64d",&n))
{
ans = ;
LL i,j,k = ;
if(n<)
{
printf("%I64d\n",n);
return ;
}
for(i = ; i<=; i++)
{
k *= ;
if(n<k)
{
ans+=(n-k/+)*i;
// printf("%I64d %I64d %I64d\n",i,ans,(n-k/10+1)*i);
break;
}
/* else if(n==k)
{
ans+=(i+1);
break;
}*/
else
{
ans+=(a[i]-a[i-])*i;
}
// printf("%d %I64d\n",i,ans);
}
printf("%I64d\n",ans);
} return ;
}
C. Vanya and Scales
Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2(exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
The first line contains two integers w, m (2 ≤ w ≤ 109, 1 ≤ m ≤ 109) — the number defining the masses of the weights and the mass of the item.
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
3 7
YES
100 99
YES
100 50
NO
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
要用质量为w0,w1,...,w100的砝码各1个称出重量m,砝码可以放在天平左边也可以放在右边。问是否可以称出,输出YES或NO。
如样例3,7:左边放3和物品,右边放1和9即可。
假设可以称出,则用w进制表示m,每一位上一定是0,1或w - 1,否则一定不行。
而如果某一位是w - 1则说明当前砝码跟物品放在一起,相当于给物品加上了这个砝码的重量。
我们只需要模拟这个过程,提取m的每一位然后计算即可。
转载请注明出处:寻找&星空の孩子
题目链接:http://codeforces.com/contest/552/problem/C
#include<string.h>
#include<stdio.h>
#define LL __int64
LL w,m;
int main()
{
scanf("%I64d%I64d",&w,&m);
if(w<=)
{
printf("YES\n");
return ;
}
while(m)
{
if(!((m-)%w)) m--;
else if(!((m+)%w)) m++;
else if(m%w) {printf("NO\n");return ;}
m=m/w;
}
printf("YES\n");
return ;
}
D. Vanya and Triangles
Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.
The first line contains integer n (1 ≤ n ≤ 2000) — the number of the points painted on the plane.
Next n lines contain two integers each xi, yi ( - 100 ≤ xi, yi ≤ 100) — the coordinates of the i-th point. It is guaranteed that no two given points coincide.
In the first line print an integer — the number of triangles with the non-zero area among the painted points.
4
0 0
1 1
2 0
2 2
3
3
0 0
1 1
2 0
1
1
1 1
0
Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).
Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).
Note to the third sample test. A single point doesn't form a single triangle.
给你二维坐标下的n个点,问一共能构成多少个面积不为0的三角形。
转载请注明出处:寻找&星空の孩子
#include<cstdio>
#include<cmath>
#include<iostream>
#define PI acos(-1.0)
using namespace std; struct Point
{
double x,y;
Point(double x=,double y=):x(x),y(y){} }; typedef Point Vector; Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);} Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);} Vector operator * (Vector A,double p){return Vector(A.x*p,A.y*p);} Vector operator / (Vector A,double p){return Vector(A.x/p,A.y/p);} bool operator < (const Point& a,const Point& b){return a.x<b.x||(a.x==b.x && a.y<b.y);} const double eps = 1e-; int dcmp(double x){if(fabs(x)<eps)return ;else return x < ? - : ;} bool operator == (const Point& a,const Point& b){return dcmp(a.x-b.x)== && dcmp(a.y-b.y)==;} double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}
double length(Vector A){return sqrt(Dot(A,A));}
double Angle(Vector A,Vector B){return acos(Dot(A,B)/length(A)/length(B));} double Cross(Vector A,Vector B){return A.x*B.y-B.x*A.y;}
double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);} inline Point read_point(Point &P)
{
scanf("%lf%lf",&P.x,&P.y);
return P;
}
int main()
{
int n;
Point po[];
scanf("%d",&n);
for(int i=;i<n;i++)
read_point(po[i]);
if(n<){printf("0\n");return ;}
int cnt=;
for(int i=;i<n;i++)
{
for(int j=i+;j<n;j++)
{
for(int k=j+;k<n;k++)
{
if(Area2(po[i],po[j],po[k])!=) cnt++;
}
}
}
printf("%d\n",cnt);
return ;
}
E. Vanya and Brackets
Vanya is doing his maths homework. He has an expression of form , where x1, x2, ..., xn are digits from 1 to 9, and sign
represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
The first line contains expression s (1 ≤ |s| ≤ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
In the first line print the maximum possible value of an expression.
3+5*7+8*4
303
2+3*5
25
3*4*5
60
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
给你一个表达式,只有乘号和加号,数字都是1到9,要求加一个括号,使得表达式的值最大,问最大是多少。乘号个数<=15。
左括号一定在乘号右边,右括号一定在乘号左边,因为如果不是这样的话,一定可以调整括号的位置使表达式的值增大。这个应该不难想。
于是只要枚举括号的位置然后计算表达式即可。
转载请注明出处:寻找&星空の孩子
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
const int N = ;
#define LL __int64 char fh[N],s[N];
LL num[N];
int ftop,ntop ,slen;
void calculate(){
if(fh[ftop]=='+')
num[ntop-]+=num[ntop],ntop--;
else if(fh[ftop]=='-')
num[ntop-]-=num[ntop],ntop--;
else if(fh[ftop]=='*')
num[ntop-]*=num[ntop],ntop--;
else if(fh[ftop]=='/')
num[ntop-]/=num[ntop],ntop--;
ftop--;
}
void countfuction(int l,int r){
ftop=;ntop=;
int flagNum=;
LL ans=;
for(int i=; i<=slen; ++i){ if(i!=slen&&(s[i]>=''&&s[i]<='')){
ans=ans*+s[i]-'';
flagNum=;
}
else{
if(flagNum)num[++ntop]=ans; flagNum=; ans=;
if(i==slen)break;
if(s[i]=='+'||s[i]=='-'){
while(ftop&&fh[ftop]!='(') calculate();
fh[++ftop]=s[i];
}
else if(s[i]=='*'&&i==r){
while(ftop&&fh[ftop]!='(') calculate(); ftop--;
while(ftop&&(fh[ftop]=='*'||fh[ftop]=='/')) calculate();
fh[++ftop]=s[i];//printf("# ");
}
else if(s[i]=='*'||i==l){
while(ftop&&(fh[ftop]=='*'||fh[ftop]=='/')) calculate();
fh[++ftop]=s[i];
if(i==l)
fh[++ftop]='(';
}
}
}
while(ftop) calculate(); }
int main(){ while(scanf("%s",s)>){
LL ans=;
int id[],k=;
for(int i=strlen(s); i>=; i--)
s[i+]=s[i];
s[]=''; s[]='*';
slen=strlen(s);
s[slen]='*'; s[slen+]=''; s[slen+]='\0';
slen=strlen(s);
for(int i=; i<slen; i++)
if(s[i]=='*')
id[k++]=i; for(int i=; i<k-; i++)
for(int j=i+; j<k; j++){
countfuction(id[i],id[j]);
if(num[]>ans)
ans=num[];
}
printf("%I64d\n",ans);
}
}
Codeforces Round #308 (Div. 2)的更多相关文章
- 水题 Codeforces Round #308 (Div. 2) A. Vanya and Table
题目传送门 /* 水题:读懂题目就能做 */ #include <cstdio> #include <iostream> #include <algorithm> ...
- 数学 Codeforces Round #308 (Div. 2) B. Vanya and Books
题目传送门 /* 水题:求总数字个数,开long long竟然莫名其妙WA了几次,也没改啥又对了:) */ #include <cstdio> #include <iostream& ...
- 暴力/进制转换 Codeforces Round #308 (Div. 2) C. Vanya and Scales
题目传送门 /* 题意:问是否能用质量为w^0,w^1,...,w^100的砝码各1个称出重量m,砝码放左边或在右边 暴力/进制转换:假设可以称出,用w进制表示,每一位是0,1,w-1.w-1表示砝码 ...
- Codeforces Round #308 (Div. 2) D. Vanya and Triangles 水题
D. Vanya and Triangles Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/55 ...
- Codeforces Round #308 (Div. 2) C. Vanya and Scales dfs
C. Vanya and Scales Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/552/p ...
- Codeforces Round #308 (Div. 2)B. Vanya and Books 数学
B. Vanya and Books Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/552/pr ...
- Codeforces Round #308 (Div. 2) A. Vanya and Table 暴力
A. Vanya and Table Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/552/pr ...
- Codeforces Round #308 (Div. 2)----C. Vanya and Scales
C. Vanya and Scales time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces Round #308 (Div. 2) A B C 水 数学
A. Vanya and Table time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
随机推荐
- 探秘JS的异步单线程
对于通常的developer(特别是那些具备并行计算/多线程背景知识的developer)来讲,js的异步处理着实称得上诡异.而这个诡异从结果上讲,是由js的“单线程”这个特性所导致的. 我曾尝试用“ ...
- H5内联视频总结
概述 之前写过h5内联视频,总结了一下当时做微信视频类h5的心得,随着工作中越来越多的接触h5,我有了更多的心得与经验,记下来供以后开发时参考,相信对其他人也有用. 内联视频的播放 内联视频需要用户主 ...
- Aseprite入门教程
因为最近在学cocos2d-x和vs搭配做手机游戏开发,想自己做一些素材,所以找到了这款软件,Aseprite v1.1.12.刚安装上时也是不懂该怎么操作,随着逐渐地摸索,对初始的使用有了一些了解. ...
- Python - IPython
1- IPython简介 HomePage:http://ipython.org/ IPython(interactive Python) provides a rich architecture f ...
- 吐血整理 20 道 Spring Boot 面试题,我经常拿来面试别人!
面试了一些人,简历上都说自己熟悉 Spring Boot, 或者说正在学习 Spring Boot,一问他们时,都只停留在简单的使用阶段,很多东西都不清楚,也让我对面试者大失所望. 下面,我给大家总结 ...
- 转转RN工程化历程
选型RN理由? 目前各大公司技术栈都是native端(android,iOS)以及H5端,然而这两大传统的开发方式都各有优缺点,下面表格简单汇总一下. - native端 web端 RN 开发效率 低 ...
- ES6的Promise
推荐一下我觉得不错关于Promise的好文章,通俗易懂 说起ES6的Promise就要提及一下JQ的$.when()方法,两者基本相同 面试的时候经常会问Promise,如果同学们能在回答Promis ...
- 【学习笔记】分类算法-k近邻算法
k-近邻算法采用测量不同特征值之间的距离来进行分类. 优点:精度高.对异常值不敏感.无数据输入假定 缺点:计算复杂度高.空间复杂度高 使用数据范围:数值型和标称型 用例子来理解k-近邻算法 电影可以按 ...
- mysql 开发进阶篇系列 50 表的数据导入(load data infile,mysqlimport )
一.概述 上篇讲到的表的数据导出(select .. into outfile 或者mysqldump),这篇继续讲表的数据导入,导入也同样有二个方法,分别是load data infile... 和 ...
- 微软新动向之Android和IOS应用 visual studio 2015 Cordova[原创]
自萨蒂亚·纳德拉(Satya Nadella)上任微软CEO以来,可谓是惊喜不断,仿佛让世界尤其是我们.net程序员心中又燃起了希望.先是免费提供 iOS 版和安卓版 Office:然后在 xbox ...