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);
}
}