Online Testing of Blue Bridge cup

For beginners, we recommend that you test the Blue Bridge cup online. Updating is not performed in order. When you are bored, you need to answer a few questions.

Http://lx.lanqiao.org/index.page

The recurrence formula of the Fibonacci series is: Fn = Fn-1 + Fn-2, where F1 = F2 = 1.
When n is large, Fn is very large. Now we want to know the remainder of Fn divided by 10007.

#include<cstdio>const int mod=10007;const int MAXN=1000000+10;int ans[MAXN];int main(){    ans[1]=ans[2]=1;    int n;    scanf("%d",&n);    for(int i=3;i<=n;i++)        ans[i]=(ans[i-1]+ans[i-2])%mod;    printf("%d\n",ans[n]);    return 0;}

Given the radius r of the circle, calculate the area of the circle.

#include<cstdio>#include<cmath>const double pi=acos(-1.0);int main(){double r;scanf("%lf",&r);printf("%.7lf\n",pi*r*r);return 0;}

Evaluate the value of 1 + 2 + 3 +... + n.

#include<cstdio>int main(){__int64 n;scanf("%I64d",&n);__int64 ans=(1+n)*n >>1;printf("%I64d\n",ans);return 0;}

Calculate a + B

#include <cstdio> int main(){    int a, b;    scanf("%d%d", &a, &b);    printf("%d", a+b);    return 0;}

Sort

#include <cstdio>#include<algorithm>using namespace std;const int MAXN=200+10;int a[MAXN];int main(){    int n;scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d",&a[i]);sort(a,a+n);printf("%d",a[0]);for(int i=1;i<n;i++)printf(" %d",a[i]);printf("\n");    return 0;}

Returns a series containing n integers. the first occurrence of integer a in the series is the nth.

#include <cstdio>const int MAXN=10000+10;int x[MAXN];int main(){int n,a;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&x[i]);scanf("%d",&a);int ans=-1;for(int i=1;i<=n;i++)if(x[i]==a){ans=i;break;}printf("%d\n",ans);return 0;}

Leap year judgment

#include <cstdio>int main(){   int n;   scanf("%d",&n);   if(n %4==0 && n%100!=0 || n%400==0)   printf("yes\n");   else   printf("no\n");    return 0;}

Given n hexadecimal positive integers, output the corresponding Octal numbers.

Train of Thought: first convert hexadecimal to 4-bit binary, and then the octal sequence can be obtained from every 3 digits of binary.

#include<cstdio>#include<cstring>const int MAXN=100000;char s[MAXN];char two[MAXN*4];int eight[MAXN*4];int main(){int T;scanf("%d",&T);while(T--){scanf("%s",s);int n=strlen(s),len=0;for(int i=0;i<n;i++){switch(s[i]){case '0':sprintf(two+len,"%s","0000");break;case '1':sprintf(two+len,"%s","0001");break;case '2':sprintf(two+len,"%s","0010");break;case '3':sprintf(two+len,"%s","0011");break;case '4':sprintf(two+len,"%s","0100");break;case '5':sprintf(two+len,"%s","0101");break;case '6':sprintf(two+len,"%s","0110");break;case '7':sprintf(two+len,"%s","0111");break;case '8':sprintf(two+len,"%s","1000");break;case '9':sprintf(two+len,"%s","1001");break;case 'A':sprintf(two+len,"%s","1010");break;case 'B':sprintf(two+len,"%s","1011");break;case 'C':sprintf(two+len,"%s","1100");break;case 'D':sprintf(two+len,"%s","1101");break;case 'E':sprintf(two+len,"%s","1110");break;case 'F':sprintf(two+len,"%s","1111");break;}len+=4;}//for(int i=0;i<len;i+=4)//printf("%c%c%c%c ",two[i],two[i+1],two[i+2],two[i+3]);int i=len-1,len2=0;for(i=len-1;i>=2;i-=3){eight[len2++]= two[i]-'0'+(two[i-1]-'0')*2+(two[i-2]-'0')*4;}if(i==2)eight[len2++]= two[i]-'0'+(two[i-1]-'0')*2+(two[i-2]-'0')*4;else if(i==1)eight[len2++]= two[i]-'0'+(two[i-1]-'0')*2;else if(i==0)eight[len2++]= two[i]-'0';i=len2-1;while(eight[i]==0)i--;for(;i>=0;i--)printf("%d",eight[i]);printf("\n");}return 0;}

Input a positive hexadecimal number string of no more than 8 digits from the keyboard, convert it to a positive decimal number, and then output it.

Note: 10 ~ 15 are represented by uppercase letters A, B, C, D, E, and F.

Idea: expand by right. Note that int is out of bounds when the range is 8 F.

#include<cstdio>#include<cstring>const int MAXN=10;char s[MAXN];int main(){while(~scanf("%s",s)){int n=strlen(s);__int64 ans=0,p=16;for(int i=0;i<n;i++){if(s[i]>='0' && s[i]<='9')ans=ans*p+(s[i]-'0');elseans=ans*p+(s[i]-'A'+10);}printf("%I64d\n",ans);}return 0;}

123321 is a very special number, which reads from the left and from the right is the same.
Enter a positive integer n to calculate all the five-and six-digit decimal numbers. The sum of the values is equal to n.

#include<cstdio>#include<cstring>char s[10];int main(){int n;scanf("%d",&n);for(int i=10000;i<1000000;i++){sprintf(s,"%d",i);int len=strlen(s);bool ok=true;for(int k=0;k<3;k++)if(s[k]!=s[len-k-1])ok=false;if(ok){int sum=0;for(int k=0;k<len;k++)sum+=s[k]-'0';if(sum==n)printf("%s\n",s);}}return 0;}

1221 is a very special number. It reads from the left and from the right is the same. It is programmed to calculate all such four-digit decimal numbers.

#include<cstdio>#include<cstring>char s[10];int main(){for(int i=1000;i<10000;i++){sprintf(s,"%d",i);int len=strlen(s);bool ok=true;for(int k=0;k<3;k++)if(s[k]!=s[len-k-1])ok=false;if(ok){printf("%s\n",s);}}return 0;}

153 is a very special number, which is equal to the Cube sum of each digit, that is, 153 = 1*1*1 + 5*5 + 3*3*3. Program all three decimal numbers that meet this condition

#include<cstdio>#include<cstring>char s[10];int main(){for(int i=100;i<1000;i++){sprintf(s,"%d",i);int sum=0;for(int k=0;k<3;k++){int x=s[k]-'0';sum+=x*x*x;}if(sum==i)printf("%d\n",i);}return 0;}

Returns n numbers to find the maximum, minimum, and sum of n numbers.

PS: the example of the question is wrong. And

#include<cstdio>const int INF=10000+10;int main(){int n,min=INF,max=-INF,sum=0,temp;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&temp);sum+=temp;if(max <temp)max=temp;if(min >temp)min=temp;}printf("%d\n%d\n%d\n",max,min,sum);return 0;}