cf327Bcf152B152C
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<list>
#include<vector>
#include<cmath>
using namespace std;
int main()
{
int n,m,i,j,k;
scanf("%d",&n);
for(i=1000001;i<=1000001+n-1;++i){
printf("%d ",i);
}
return 0;
}
One day Vasya went out for a walk in the yard but there weren't any of his friends outside and he had no one to play touch and run. But the boy didn't lose the high spirits and decided to play touch and run with himself. You may ask: "How did he do that?" The answer is simple.
Vasya noticed that the yard is a rectangular n × m field. The squares have coordinates (x, y) (1 ≤ x ≤ n, 1 ≤ y ≤ m), where x is the index of the row and y is the index of the column.
Initially Vasya stands in the square with coordinates (xc, yc). To play, he has got a list of k vectors (dxi, dyi) of non-zero length. The game goes like this. The boy considers all vectors in the order from 1 to k, and consecutively chooses each vector as the current one. After the boy has chosen a current vector, he makes the maximally possible number of valid steps in the vector's direction (it is possible that he makes zero steps).
A step is defined as one movement from the square where the boy is standing now, in the direction of the current vector. That is, if Vasya is positioned in square (x, y), and the current vector is (dx, dy), one step moves Vasya to square (x + dx, y + dy). A step is considered valid, if the boy does not go out of the yard if he performs the step.
Vasya stepped on and on, on and on until he ran out of vectors in his list. Ha had been stepping for so long that he completely forgot how many steps he had made. Help the boy and count how many steps he had made.
The first input line contains two integers n and m (1 ≤ n, m ≤ 109) — the yard's sizes. The second line contains integers xc and yc — the initial square's coordinates (1 ≤ xc ≤ n, 1 ≤ yc ≤ m).
The third line contains an integer k (1 ≤ k ≤ 104) — the number of vectors. Then follow k lines, each of them contains two integers dxiand dyi (|dxi|, |dyi| ≤ 109, |dx| + |dy| ≥ 1).
Print the single number — the number of steps Vasya had made.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the%I64d specificator.
4 5 1 1 3 1 1 1 1 0 -2
4
10 10 1 2 1 -1 0
0
In the first sample Vasya is initially positioned at square (1, 1) and makes 3 steps by the first vector (1, 1). So, he consecutively visits the squares (2, 2), (3, 3), (4, 4). Then he makes 0 steps by the second vector (1, 1). He makes 1 more step by the third vector(0, - 2) and he ends up in square (4, 2). Overall, Vasya makes 4 steps.
In the second sample Vasya is initially positioned in square (1, 2) and makes 0 steps by vector ( - 1, 0), as the square with coordinates(0, 2) is located outside the yard.
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
long long n,m,xc,yc,k,ans=0,x,y;
scanf("%lld%lld%lld%lld%lld",&n,&m,&xc,&yc,&k);
for(int i=0;i<k;++i){
scanf("%lld%lld",&x,&y);
long long tempx=inf,tempy=inf,temp;
if(x>0){
tempx=(n-xc)/x;
}
else if(x<0){
tempx=abs((xc-1)/x);
}
if(y>0){
tempy=(m-yc)/y;
}
else if(y<0){
tempy=abs((yc-1)/y);
}
temp=min(tempx,tempy);
ans+=temp;xc=xc+temp*x;yc=yc+temp*y;
}
printf("%lld\n",ans);
return 0;
}
One day little Vasya found mom's pocket book. The book had n names of her friends and unusually enough, each name was exactly mletters long. Let's number the names from 1 to n in the order in which they are written.
As mom wasn't home, Vasya decided to play with names: he chose three integers i, j, k (1 ≤ i < j ≤ n, 1 ≤ k ≤ m), then he took names number i and j and swapped their prefixes of length k. For example, if we take names "CBDAD" and "AABRD" and swap their prefixes with the length of 3, the result will be names "AABAD" and "CBDRD".
You wonder how many different names Vasya can write instead of name number 1, if Vasya is allowed to perform any number of the described actions. As Vasya performs each action, he chooses numbers i, j, k independently from the previous moves and his choice is based entirely on his will. The sought number can be very large, so you should only find it modulo 1000000007 (109 + 7).
The first input line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of names and the length of each name, correspondingly. Then n lines contain names, each name consists of exactly m uppercase Latin letters.
Print the single number — the number of different names that could end up in position number 1 in the pocket book after the applying the procedures described above. Print the number modulo 1000000007 (109 + 7).
2 3 AAB BAA
4
4 5 ABABA BCGDG AAAAA YABSA
216
In the first sample Vasya can get the following names in the position number 1: "AAB", "AAA", "BAA" and "BAB".
统计出每一列中不同的字符的个数相乘即可
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
char str[110][110];
bool vis[110][30];
int main()
{
int n,m,i,j,k;
scanf("%d%d",&n,&m);
for(i=1;i<=n;++i){
scanf("%s",str[i]+1);
}
long long ans=1;
for(i=1;i<=m;++i){
long long temp=0;
for(j=1;j<=n;++j){
if(!vis[i][str[j][i]-'A']){
temp++;vis[i][str[j][i]-'A']=true;
}
}
ans*=temp;ans%=1000000007;
}
printf("%lld\n",ans);
return 0;
}