牛牛养了n只奶牛,牛牛想给每只奶牛编号,这样就可以轻而易举地分辨它们了。 每个奶牛对于数字都有自己的喜好,第i只奶牛想要一个1和x[i]之间的整数(其中包含1和x[i])。
牛牛需要满足所有奶牛的喜好,请帮助牛牛计算牛牛有多少种给奶牛编号的方法,输出符合要求的编号方法总数。
输入包括两行,第一行一个整数n(1 ≤ n ≤ 50),表示奶牛的数量 第二行为n个整数x[i](1 ≤ x[i] ≤ 1000)
输出一个整数,表示牛牛在满足所有奶牛的喜好上编号的方法数。因为答案可能很大,输出方法数对1,000,000,007的模。
4 4 4 4 4
24
#include <iostream>#include <algorithm>using namespace std;long long mod=1000000007;int main(){int n;while( cin >> n ){int a[n];for(int i=0; i<n; i++)cin >> a[i];sort(a,a+n);long long counts=1;for(int i=0; i<n; i++){counts=((a[i]-i)*counts)%mod;}cout << counts << endl;}}
本套3道题的C++代码已挂到了我的GitHub(https://github.com/shiqitao/NowCoder-Solutions),牛客网上的其他题目解答也在持续更新。
#include <iostream> #include <algorithm> #define mod 1000000007 using namespace std; int main() { int n; cin >> n; int *num = new int[n]; for (int i = 0; i < n; i++) { cin >> num[i]; } sort(num, num + n); long long int result = 1; for (int i = 0; i < n; i++) { if (num[i] <= i) { cout << 0 << endl; return 0; } result = result*(num[i] - i) % mod; } cout << result << endl; delete[] num; return 0; }
import java.util.Arrays; import java.util.Scanner; public class Main { public static final long MAX = 1000000007; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); long[] arr = new long[n]; for (int i = 0; i < n; i++) { arr[i] = sc.nextInt(); } Arrays.sort(arr); long res = 1; for (int i = 0; i < n; i++) { res *= (arr[i] - i) % MAX; res %= MAX; } System.out.println(res); } }
import java.util.*; public class Main{ public static void main(String[] args){ try(Scanner in = new Scanner(System.in)){ int n = in.nextInt(),i = 0; int[] a = new int[n]; while(i < n){ a[i++] = in.nextInt(); } System.out.println(helper(a)); } } public static long helper(int[] a){ long res = 1; Arrays.sort(a); int c = 0; for(int num:a){ res = (res * ((num - c++) % 1000000007)) % 1000000007; //先取余再乘再取余 不用等最后结果再取余 } return res; } }
#include<iostream>
#include<algorithm>
using namespace std;
long long mod = 1000000007;
int main() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
long long counter = 1;
for (int i = 0; i < n; i++) {
counter = ((a[i] - i)*counter)%mod;
}
cout << counter << endl;
return 0;
}
import java.util.*; public class Main { private static final long MOD = 1000000007; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int[] nums = new int[n]; for (int i=0; i!=n; i++) { nums[i] = sc.nextInt(); } Arrays.sort(nums); long ans = 1; for (int i=0; i!=n; i++) { ans *= ((nums[i] - i) % MOD); ans %= MOD; } System.out.println(ans); } }
import java.math.BigInteger; import java.util.*; public class TheCowNumber { static final int mod = 1000000007; public static void main(String[] args) { /** * 本题重点在于如何处理大数。而不是算法。 * 算法很简单,就是排序,每选一个数,后一个可选择的就减一 */ Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); Integer array[] = new Integer[n]; for(int i=0;i < n;i ++){ array[i] = scanner.nextInt(); } Arrays.sort(array); BigInteger result = new BigInteger("1"); for(int i=0;i <array.length;i ++){ int tmp = array[i] - i; String s = tmp + ""; result = result.multiply(new BigInteger(s)); } String mod = 1000000007 + ""; System.out.println(result.mod(new BigInteger(mod))); } }
#include<iostream> #include<vector> #include<algorithm> using namespace std; int main() { int N; vector<int> X; cin >> N; for(int i = 0; i < N; i++){ int temp; cin >> temp; X.push_back(temp); } sort(X.begin(), X.end()); for(int i=0; i < N; i++){ X[i] -= i; } long long product = 1; for(int i = 0; i < N; i++){ product = product * X[i] % 1000000007; } cout << product; return 0; }
import sys n=int(sys.stdin.readline().strip()) x=list(map(int,sys.stdin.readline().split())) x.sort() res=1 for i in range(len(x)): res=res*(x[i]-i) print(res%(1000000007))
def solution(): n = int(input()) line = list(map(int, input().strip().split())) if not line:return None line.sort() res = 1 for i in range(n): tmp = line[i] - i res *= tmp return res % 1000000007 if __name__ == '__main__': s = solution() print(s)
import java.util.*; public class Main{ public static void main(String[] args){ Scanner sc=new Scanner(System.in); while(sc.hasNext()){ CAL(sc); } } public static void CAL(Scanner sc){ int n=sc.nextInt(); int[] x=new int[n]; for (int i=0;i<n;i++){ x[i]=sc.nextInt(); } Arrays.sort(x); long R=1; for(int i=0;i<n;i++){ R=(R%1000000007)*(x[i]-i);//这里直接取余再接着乘,不要等最后结果再乘 } System.out.println(R); } }
n = int(input()) x = list(map(int, input().split())) x.sort() out = x[0] for i in range(1,len(x)): out = out * (x[i]-i) print(out%1000000007)
#include<bits/stdc++.h> using namespace std; int main() { int n,a; vector<int>v; cin>>n; for(int i=0;i<n;i++) { cin>>a; v.push_back(a); } sort(v.begin(),v.end()); long long sum=1,mod=1000000007; for(int i=0;i<n;i++) { sum=sum%mod*(v[i]-i)%mod; } cout<<sum; return 0; }