The bored BaoBao is playing a number game. In the beginning, there are numbers. For each turn, BaoBao will take out two numbers from the remaining numbers, and calculate the product of them.
Now, BaoBao is curious to know the minimum sum of the products if he plays at least turns. Can you tell him?
Input
The first line of input contains a positive integer (about 30), the number of test cases. For each test case:
The first line contains two integers and (, ). Their meanings are described above.
The second line contains integers (), indicating the numbers.
Output
For each test case output one integer, indicating the minimum sum of the products.
Sample Input
3 4 2 1 3 2 4 3 1 2 3 1 4 0 1 3 2 4
Sample Output
10 2 0
Hint
For the first sample test case, the answer is 1 × 4 + 3 × 2 = 10.
For the second sample test case, the answer is 2 × 1 = 2.
n个数找m组数，使乘积的和最小
让前m小最大和最小的相乘就好了
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int a[N]; int main() { int T; cin>>T; while(T--) { int n,m; cin>>n>>m; for(int i=0; i<n; i++) cin>>a[i]; sort(a,a+n); int r=2*m-1; long long s=0; for(int i=0; i<m; i++) s+=a[i]*a[r-i]; cout<<s<<endl; } }
Super Brain is a famous scientific reality and talent show aiming to find people with exceptional brainpower.
In one of the challenges, two integer sequences and of length are given to the contestant. It's guaranteed that and hold for all , and there is exactly one integer in the first sequence which also appears in the second sequence. The contestant has to memorize the two sequences in a very short time, and find the integer which appears in both sequences correctly.
As a technical staff of the show, you are required to write a program and find out the correct integer.
Input
There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:
The first line contains an integer (), indicating the length of the sequence.
The second line contains integers (), indicating the first sequence.
The third line contains integers (), indicating the second sequence.
It's guaranteed that and hold for all , and there is exactly one integer in the first sequence which also appears in the second sequence.
It's also guaranteed that the sum of over all test cases will not exceed .
Output
For each test case output one line containing one integer, indicating the integer which appears in both sequences.
Sample Input
3 3 3 1 2 5 3 4 2 38324 14122 38323 14122 1 180310 180310
Sample Output
3 14122 180310
我可能被卡常了，memset就过了
找到两个数组唯一相同的
#include<bits/stdc++.h> using namespace std; const int N=1e6+5; int a[N],n; int main() { ios::sync_with_stdio(false); int T; cin>>T; while(T--) { memset(a,0,sizeof a); cin>>n; int ans; for(int i=0,x; i<n; i++)cin>>x,a[x]++; for(int i=0,x; i<n; i++) { cin>>x; if(a[x])ans=x; } cout<<ans<<" "; } }
A sequence of integers () is called a happy sequence if each number divides (without a remainder) the next number in the sequence. More formally, we can say for all , or we can say for all .
Given and , find the number of happy sequences of length . Two sequences and are different, if and only if there exists an such that and .
As the answer can be rather large print it modulo ().
Input
There are multiple test cases. The first line of the input contains an integer (about 50), indicating the number of test cases. For each test case:
The first and only line contains two integers and (), indicating the upper limit of the elements in the sequence and the length of the sequence.
Output
For each case output a single integer, indicating the number of happy sequences of length modulo
Sample Input
1 3 2
Sample Output
5
Hint
In the sample test case, the happy sequences are: , , , , .
一个数组是按照倍数关系的
这个题利用埃筛的思想去枚举i个数s*j作为最后一个元素就好
using namespace std; typedef long long ll; const ll MD=1e9+7; const int N=2005; ll dp[N][N]; int main() { ios::sync_with_stdio(false); int T; cin>>T; while(T--) { int n,k; cin>>n>>k; memset(dp,0,sizeof dp); for(int i=1; i<N; i++) dp[1][i]=1; for(int i=1; i<k; i++) for(int j=1; j<=n; j++) for(int s=1; s*j<=n; s++) dp[i+1][s*j]=(dp[i+1][s*j]+dp[i][j])%MD; ll sum=0; for(int i=1; i<=n; i++) sum=(sum+dp[k][i])%MD; cout<<(sum+MD)%MD<<endl; } return 0; }
BaoBao is traveling along a line with infinite length.
At the beginning of his trip, he is standing at position 0. At the beginning of each second, if he is standing at position , with probability he will move to position , with probability he will move to position , and with probability he will stay at position . Positions can be positive, 0, or negative.
DreamGrid, BaoBao's best friend, is waiting for him at position . BaoBao would like to meet DreamGrid at position after exactly seconds. Please help BaoBao calculate the probability he can get to position after exactly seconds.
It's easy to show that the answer can be represented as , where and are coprime integers, and is not divisible by . Please print the value of modulo , where is the multiplicative inverse of modulo .
Input
There are multiple test cases. The first line of the input contains an integer (about 10), indicating the number of test cases. For each test case:
The first and only line contains two integers and (). Their meanings are described above.
Output
For each test case output one integer, indicating the answer.
Sample Input
3 2 -2 0 0 0 1
Sample Output
562500004 1 0
1s分为2s，只要n+m挑对了就行了
#include <bits/stdc++.h> using namespace std; typedef long long ll; const ll MD=1e9+7; const int N=2e5+5; ll la(ll a,ll b) { a%=MD; ll ans=1; while(b>0) { if(b&1)ans=ans*a%MD; b>>=1; a=a*a%MD; } return ans; } ll f[N],v[N],ans; ll C(ll n,ll m) { if(m<0||m>n) return 0; return f[n]*v[m]%MD*v[n-m]%MD; } int main() { ios::sync_with_stdio(false); f[0]=1; for (ll i=1; i<N; i++) f[i]=f[i-1]*i%MD; v[N-1]=la(f[N-1],MD-2); for (ll i=N-2; i>=0; i--) v[i]=v[i+1]*(i+1LL)%MD; int T; cin>>T; while(T--) { int n,m; cin>>n>>m; ans=C(2*n,m+n)*la(la(2,2*n),MD-2)%MD; cout<<ans<<" "; } return 0; }
BaoBao is keen on collection. Recently he is abandoning himself to Kantai Collection, especially to collecting cute girls, known as "Fleet Girls".
There are various types of girls in the game. To get a girl, one can use some materials to build her. The probability to get a type of girl by building is the same for all types of girls. From the Coupon Collector's Problem we know that, to collect all types of girls, the expected number of times of building is .
But this rule does not apply to BaoBao, as he is always luckier than the ordinary players (maybe because he's an European). For BaoBao to collect all types of girls, the expected number of times of building is , where means the maximum integer that doesn't exceed .
As a lucky man, BaoBao is not interested in the actual value of , and he just wants to know whether is odd or even. Can you help him?
Input
The first line of the input is an interger (about 100), indicating the number of test cases. For each test case:
The first line contains an integer (), indicating the number of types of girls.
Output
For each test case, if is odd output "1" (without quotes), if is even output "0" (without quotes).
Sample Input
9 2 3 23 233 2333 23333 233333 2333333 23333333
Sample Output
1 1 0 1 0 0 1 1 0
被卡了一波时间，开方啊
def sqrt(n): l=0 r=n while l<=r : mi=(l+r)>> 1 F=mi*mi if(F<n): l=mi+1 elif(F>n): r=mi-1 else: return mi return l-1 T=int(input()) for i in range(0,T): a=int(input()) if a==0: print("0") elif a<=3: print("1") else: a=sqrt(a) print(a & 1)
sqrt对于每个都是成立的
def sqrt(n): l=0 r=n while l<=r : mi=(l+r)>> 1 F=mi*mi if(F<n): l=mi+1 elif(F>n): r=mi-1 else: return mi return l-1 T=int(input()) for i in range(0,T): a=int(input()) a=sqrt(a) print(a & 1)