UVa 12712 && UVaLive 6653 Pattern Locker (排列组合)

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
", kase, ans);
    }
    return 0;
}