Floor problem
In this problem, we have f(n,x)=Floor[n/x]. Here Floor[x] is the biggest integer such that no larger than x. For example, Floor[1.1]=Floor[1.9]=1, Floor[2.0]=2.
You are given 3 positive integers n, L and R. Print the result of f(n,L)+f(n,L+1)+...+f(n,R), please.
Input
The first line of the input contains an integer T (T≤100), indicating the number of test cases.
Then T cases, for any case, only 3 integers n, L and R (1≤n, L, R≤10,000, L≤R).
Output
For each test case, print the result of f(n,L)+f(n,L+1)+...+f(n,R) in a single line.
Sample Input
3 1 2 3 100 2 100 100 3 100
Sample Output
0 382 332
水题
#include<cstdio> #include<cstring> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<iostream> #include<algorithm> #include<cmath> const int maxn=1e5+5; typedef long long ll; using namespace std; int a[maxn]; int main() { int T; cin>>T; while(T--) { int n,L,R; scanf("%d%d%d",&n,&L,&R); ll sum=0; for(int t=L;t<=R;t++) { sum+=(n/t); } printf("%lld ",sum); } return 0; }
Solve equation
You are given two positive integers A and B in Base C. For the equation:
We know there always existing many non-negative pairs (k, d) that satisfy the equation above. Now in this problem, we want to maximize k.
For example, A="123" and B="100", C=10. So both A and B are in Base 10. Then we have:
(1) A=0*B+123
(2) A=1*B+23
As we want to maximize k, we finally get one solution: (1, 23)
The range of C is between 2 and 16, and we use 'a', 'b', 'c', 'd', 'e', 'f' to represent 10, 11, 12, 13, 14, 15, respectively.
Input
The first line of the input contains an integer T (T≤10), indicating the number of test cases.
Then T cases, for any case, only 3 positive integers A, B and C (2≤C≤16) in a single line. You can assume that in Base 10, both A and B is less than 2^31.
Output
For each test case, output the solution “(k,d)” to the equation in Base 10.
Sample Input
3 2bc 33f 16 123 100 10 1 1 2
Sample Output
(0,700) (1,23) (1,0)
代码：
#include<cstdio> #include<cstring> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<iostream> #include<algorithm> #include<cmath> #include<string> using namespace std; char str1[1005],str2[1005]; int power(int x,int y) { int ans=1; for(int t=0;t<y;t++) { ans=ans*x; } return ans; } int main() { int T; cin>>T; int s1,s2,s3; int ss; while(T--) { s1=0; s2=0; s3=0; cin>>str1>>str2; scanf("%d",&ss); int len1=strlen(str1); int len2=strlen(str2); for(int t=len1-1;t>=0;t--) { if(str1[t]>='0'&&str1[t]<='9') s1+=(str1[t]-'0')*power(ss,len1-1-t); else { s1+=(str1[t]-'a'+10)*power(ss,len1-1-t); } } for(int t=len2-1;t>=0;t--) { if(str2[t]>='0'&&str2[t]<='9') s2+=(str2[t]-'0')*power(ss,len2-1-t); else { s2+=(str2[t]-'a'+10)*power(ss,len2-1-t); } } int k,d; if(s2==0) { k=s1; d=s1; } else { k=(s1/s2); d=(s1-(k)*s2)%s2; } printf("(%d,%d) ",k,d); } return 0; }
Star
Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sky. Suddenly, one of Overpower’s classmates ask him: “How many acute triangles whose inner angles are less than 90 degrees (regarding stars as points) can be found? Assuming all the stars are in the same plane”. Please help him to solve this problem.
Input
The first line of the input contains an integer T (T≤10), indicating the number of test cases.
For each test case:
The first line contains one integer n (1≤n≤100), the number of stars.
The next n lines each contains two integers x and y (0≤|x|, |y|≤1,000,000) indicate the points, all the points are distinct
Output
For each test case, output an integer indicating the total number of different acute triangles.
Sample Input
1 3 0 0 10 0 5 1000
Sample Output
1
代码：
#include<cstdio> #include<cstring> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<iostream> #include<algorithm> #include<cmath> #include<string> using namespace std; struct node { double x; double y; }p[105]; int main() { int T; cin>>T; while(T--) { int n; scanf("%d",&n); for(int t=0;t<n;t++) { scanf("%lf%lf",&p[t].x,&p[t].y); } int s=0; for(int t=0;t<n;t++) { for(int j=t+1;j<n;j++) { for(int k=j+1;k<n;k++) { double a=(p[t].x-p[j].x)*(p[t].x-p[j].x)+(p[t].y-p[j].y)*(p[t].y-p[j].y); double b=(p[t].x-p[k].x)*(p[t].x-p[k].x)+(p[t].y-p[k].y)*(p[t].y-p[k].y); double c=(p[j].x-p[k].x)*(p[j].x-p[k].x)+(p[j].y-p[k].y)*(p[j].y-p[k].y); if(a>=b&&a>=c) { if(b+c-a>0) { s++; } } else if(b>=a&&b>=c) { if(a+c-b>0) { s++; } } else if(c>=a&&c>=b) { if(a+b-c>0) { s++; } } } } } printf("%d ",s); } return 0; }
Min Number
Now you are given one non-negative integer n in 10-base notation, it will only contain digits ('0'-'9'). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].
For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.
Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?
Please note that in this problem, leading zero is not allowed!
Input
The first line of the input contains an integer T (T≤100), indicating the number of test cases.
Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.
Output
For each test case, output the minimum number we can get after no more than M operations.
Sample Input
3 9012 0 9012 1 9012 2
Sample Output
9012 1092 1029
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<cmath> using namespace std; const int maxn=1e5+5; typedef long long ll; char str[maxn]; char a[maxn]; int main() { int T; cin>>T; int m; int len; while(T--) { scanf("%s%d",str,&m); len=strlen(str); int s=0; for(int t=0;t<len;t++) { if(s==m) { break; } char minn=str[t]; int k=t; for(int j=t;j<len;j++) { if(t==0) { if(minn>str[j]&&str[j]!='0') { minn=str[j]; k=j; } } else { if(minn>str[j]) { minn=str[j]; k=j; } } } if(k!=t) { swap(str[t],str[k]); s++; } } printf("%s ",str); } return 0; }