UVa 12712 && UVaLive 6653 Pattern Locker (排列组合)
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题意:给定 一个n * n 的宫格,就是图案解锁,然后问你在区间 [l, r] 内的所有的个数进行组合,有多少种。
析:本来以为是数位DP,后来仔细一想是排列组合,因为怎么组合都行,不用考虑实际要考虑的比如 要连13,必须经过2,这个可以不用。
所以这题就是A(n,m)。剩下的就简单了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 10000 + 5; const LL mod = 10000000000007; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } LL sum[maxn]; int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ int k; scanf("%d %d %d", &n, &m, &k); n = n * n; LL ans = 0; m = n - m + 1; k = n - k + 1; sum[n] = n; for(int i = n-1; i >= 1; --i) sum[i] = (sum[i+1] * i) % mod; for(int i = k; i <= m; ++i) ans = (ans + sum[i]) % mod; printf("Case %d: %lld\n", kase, ans); } return 0; }
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