題目鏈接：https://code.google.com/codejam/contest/3324486/dashboard#s=p2

題目：

思路：

　　

?

代碼實現(xiàn)如下：

?

1 #include <set> 2 #include <map> 3 #include <deque> 4 #include <queue> 5 #include <stack> 6 #include <cmath> 7 #include <ctime> 8 #include <bitset> 9 #include <cstdio> 10 #include <string> 11 #include <vector> 12 #include <cstdlib> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 18 typedef long long LL; 19 typedef pair<LL, LL> pLL; 20 typedef pair<LL, int> pli; 21 typedef pair<int, LL> pil;; 22 typedef pair<int, int> pii; 23 typedef unsigned long long uLL; 24 25 #define lson rt<<1 26 #define rson rt<<1|1 27 #define lowbit(x) x&(-x) 28 #define name2str(name) (#name) 29 #define bug printf("*********\n") 30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl 31 #define FIN freopen("D://code//in.txt", "r", stdin) 32 #define IO ios::sync_with_stdio(false),cin.tie(0) 33 34 const double eps = 1e-8; 35 const int mod = 1000000007; 36 const int maxn = 2e5 + 7; 37 const double pi = acos(-1); 38 const int inf = 0x3f3f3f3f; 39 const LL INF = 0x3f3f3f3f3f3f3f3fLL; 40 41 int t, n, m; 42 LL A[maxn], inv[maxn]; 43 44 LL Mod_Pow(LL x, int n) { 45 LL res = 1; 46 while(n) { 47 if(n & 1) res = res * x % mod; 48 x = x * x % mod; 49 n >>= 1; 50 } 51 return res; 52 } 53 54 void init() { 55 A[0] = 1; 56 for(int i = 1; i < maxn; i++) A[i] = (A[i-1] * i) % mod; 57 inv[maxn-1] = Mod_Pow(A[maxn-1], mod - 2); 58 for(int i = maxn - 2; i >= 0; i--) { 59 inv[i] = inv[i+1] * (i + 1) % mod; 60 } 61 } 62 63 int Com(int n, int m) { 64 if(n < 0 || m < 0 || m > n) return 0; 65 if(m == 0 || m == n) return 1; 66 return ((A[n] * inv[n-m] % mod) * inv[m]) % mod; 67 } 68 69 int main() { 70 #ifndef ONLINE_JUDGE 71 FIN; 72 #endif 73 init(); 74 int icase = 0; 75 scanf("%d", &t); 76 while(t--) { 77 scanf("%d%d", &n, &m); 78 LL sum = 0; 79 for(int i = 0; i <= m; i++) { 80 if(i % 2 == 0) { 81 sum = (sum + ((Com(m, i) * Mod_Pow(2, i) % mod) * A[2 * n - i]) % mod) % mod; 82 } else { 83 sum = (sum - ((Com(m, i) * Mod_Pow(2, i) % mod) * A[2 * n - i]) % mod + mod) % mod; 84 } 85 } 86 printf("Case #%d: %lld\n", ++icase, sum % mod); 87 } 88 return 0; 89 }

?

轉(zhuǎn)載于:https://www.cnblogs.com/Dillonh/p/9978852.html

《新程序員》：云原生和全面數(shù)字化實踐50位技術(shù)專家共同創(chuàng)作，文字、視頻、音頻交互閱讀

總結(jié)

以上是生活随笔為你收集整理的Let Me Count The Ways(Kickstart Round H 2018)的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。