Iroha and a Grid
题目描述
We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell.
However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells.
Find the number of ways she can travel to the bottom-right cell.
Since this number can be extremely large, print the number modulo 109+7.
Constraints
1≤H,W≤100,000
1≤A<H
1≤B<W
However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells.
Find the number of ways she can travel to the bottom-right cell.
Since this number can be extremely large, print the number modulo 109+7.
Constraints
1≤H,W≤100,000
1≤A<H
1≤B<W
输入
The input is given from Standard Input in the following format:
H W A B
H W A B
输出
Print the number of ways she can travel to the bottom-right cell, modulo 109+7.
样例输入
2 3 1 1
样例输出
2
提示
We have a 2×3 grid, but entering the bottom-left cell is forbidden. The number of ways to travel is two: "Right, Right, Down" and "Right, Down, Right".
来源
题意:n*m矩阵 左下方A*B为禁区,每次可以走下或者左,n,m<=1e5,问从左上到右下不经过禁区时的方案数?
题解:就是找路吧 以为可以动态规划但是数据太大 数组开不了 看了下大佬的 原来可以用组合数 若无禁区,则方案数为C(n+m-2,n-1)
有禁区时 每个合法路径都会到达(n-A,i)i>B 即n-A行的某一列上.则每个合法路径都可以分成两段,(1,1)->(n-A,i),(n-A,i)->(n,m) (i>B)
注意枚举到(n-A,i)不能向右移动,否则重复枚举.所以路径变为三段.
这个题目因为是排列组合涉及到除法,所以需要使用逆元。
链接 逆元
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <cstdlib> #include <iomanip> #include <cmath> #include <ctime> #include <map> #include <set> #include <queue> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x<y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define INF 0x3f3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; ll gcd(ll a,ll b){ return b?gcd(b,a%b):a; } bool cmp(int x,int y) { return x>y; } const int N=2e5+20; const ll mod=1e9+7; ll f[N],n,m,A,B; ll powmod(ll x,ll n) { ll s=1; while(n){ if(n&1) s=(s*x)%mod; x=(x*x)%mod; n>>=1; } return s; } ll C(ll n,ll m) { ll a=f[n]; ll b=(f[m]*f[n-m])%mod; return (a*powmod(b,mod-2))%mod; } int main() { f[0]=1; for(ll i=1;i<N;i++) f[i]=(f[i-1]*i)%mod; while(cin>>n>>m>>A>>B){ ll res=0; for(ll i=B+1;i<=m;i++){ ll tmp=(C(i-1+n-A-1,n-A-1)*C(m-i+A-1,m-i))%mod; res=(res+tmp)%mod; } cout<<res<<endl; } return 0; }