Easy Number Game
Time Limit: 2 Seconds Memory Limit: 65536 KB
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
Time Limit: 1 Second Memory Limit: 65536 KB
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<<"\n";
}
}
Happy Sequence
Time Limit: 3 Seconds Memory Limit: 65536 KB
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;
}
Travel along the Line
Time Limit: 1 Second Memory Limit: 65536 KB
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<<"\n";
}
return 0;
}
原文地址：https://www.cnblogs.com/BobHuang/p/8541184.html