Gym - 100625F Count Ways 快速幂+容斥原理

题意:n*m的格子，中间有若干点不能走，问从左上角到右下角有多少种走法。

思路：CountWay(i,j) 表示从 i 点到 j 点的种数。然后用容斥原理加加减减解决

 #pragma comment(linker, "/STACK:1000000000")
 #include <iostream>
 #include <cstdio>
 #include <fstream>
 #include <algorithm>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <string>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <set>
 #define LL long long
 #define MAXN 100005
 #define MOD 1000000007
 #define INF 0x3f3f3f3f
 #define eps 1e-8
 using namespace std;
 struct Node
 {
     LL x, y;
 };
 Node p[MAXN];
 LL factor[ * MAXN], w[ * MAXN];
 LL res[MAXN];
 bool compare(Node a, Node b)
 {
     return a.x < b.x || (a.x == b.x && a.y < b.y);
 }
 LL CountWay(LL x, LL y)
 {
     LL res = factor[x + y];
     res = res * w[x] % MOD;
     res = res * w[y] % MOD;
     return res;
 }
 LL quick_power(LL x, LL y)
 {
     if (y == ){
         return (LL);
     }
     if (y == ){
         return x % MOD;
     }
     LL res = quick_power(x, y >> );
     res = (res * res) % MOD;
     if (y & ){
         res = (res * x) % MOD;
     }
     return res;
 }
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);
 #endif // OPEN_FILE
     int T;
     factor[] = ;
     w[] = ;
     for (LL i = ; i <= ; i++){
         factor[i] = (factor[i - ] * i) % MOD;
         w[i] = quick_power(factor[i], MOD - );
     }
     scanf("%d", &T);
     LL n, m, k;
     while (T--){
         scanf("%I64d%I64d%I64d", &n, &m, &k);
         for (int i = ; i <= k; i++){
             scanf("%I64d%I64d", &p[i].x, &p[i].y);
         }
         p[k + ].x = n;
         p[k + ].y = m;
         sort(p + , p + k + , compare);
         for (int i = ; i <= k + ; i++){
             res[i] = CountWay(p[i].x - , p[i].y - );
         }
         for (int i = ; i <= k + ; i++){
             for (int j = i + ; j <= k + ; j++){
                 if (p[j].x < p[i].x || p[j].y < p[i].y) continue;
                 //if(p[j].x == p[i].x && p[j].y == p[i].y) continue;
                 res[j] = (res[j] - (res[i] * CountWay(p[j].x - p[i].x, p[j].y - p[i].y)) % MOD) % MOD;
             }
         }
         printf("%I64d\n", (res[k + ] + MOD) % MOD);
     }
 }