C++-蓝桥杯-入门训练

Fibonacci数列,快速幂

#include <cstdio>
struct Matrix{int a[5][5];};
const int N=2,MOD=10007;
Matrix A,B,O,I;
Matrix Mul(Matrix A,Matrix B){
    Matrix C=O;
    for(int i=1;i<=N;i++)
        for(int j=1;j<=N;j++)
            for(int k=1;k<=N;k++)
                C.a[i][j]=(C.a[i][j]+A.a[i][k]*B.a[k][j])%MOD;
    return C;
}
Matrix Pow(Matrix A,int n){
    Matrix B=I;
    for(;n;n>>=1,A=Mul(A,A))if(n&1)B=Mul(B,A);
    return B;
}
int main(){
    for(int i=1;i<=2;i++)for(int j=1;j<=2;j++)O.a[i][j]=0,I.a[i][j]=(i==j);
    A.a[1][1]=0,A.a[1][2]=1;
    A.a[2][1]=1,A.a[2][2]=1;
    int n;scanf("%d",&n),B=Pow(A,n+1),printf("%d\n",B.a[1][1]); 
    return 0;
}

 

圆的面积

#include <cmath>
#include <cstdio>
typedef double db;
db r,pi=acos(-1);
int main(){
    scanf("%lf",&r),printf("%.7lf",pi*r*r);
    return 0;
}

 

序列求和

#include <cstdio>
typedef long long ll;
ll ans,n;
int main(){
    scanf("%lld",&n),printf("%lld",(1+n)*n/2);
    return 0;
}

 

A+B问题

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

 

数列排序，STL使用

#include <queue>
#include <cstdio>
using namespace std;
priority_queue<int>q;
int main(){
    int n,a;scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a),q.push(-a);
    for(int i=1;i<=n;i++)printf("%d ",-q.top()),q.pop();
    return 0;
}

 

十六进制转八进制，进制转化

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
using namespace std;
char s[100010];
int main(){
    int n,len,num,ok;scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%s",s+1),len=strlen(s+1);
        string ans="";
        for(int i=1;i<=len;i++)
            switch(s[i]){
                case '0':ans+="0000";break;case '1':ans+="0001";break;
                case '2':ans+="0010";break;case '3':ans+="0011";break;
                case '4':ans+="0100";break;case '5':ans+="0101";break;
                case '6':ans+="0110";break;case '7':ans+="0111";break;
                case '8':ans+="1000";break;case '9':ans+="1001";break;
                case 'A':ans+="1010";break;case 'B':ans+="1011";break;
                case 'C':ans+="1100";break;case 'D':ans+="1101";break;
                case 'E':ans+="1110";break;case 'F':ans+="1111";break;
            }//printf("%s\n",ans.c_str());            
        switch(len%3){
            case 1:ans="00"+ans;break;
            case 2:ans="0"+ans;break;
        }
        len=ans.length();ok=0;
        for(int i=0;i<len;i+=3){
            num=4*(ans[i]-'0')+2*(ans[i+1]-'0')+(ans[i+2]-'0');//printf("%c %c %c\n",ans[i],ans[i+1],ans[i+2]);
            if(num)ok=1;if(ok)putchar(num+'0');
        }
        puts("");
    }
    return 0;
}

 

阶乘计算，高精度

#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
using namespace std;
const int MAXN=5000;
char buf[MAXN];
struct BigData{
    int a[MAXN],cnt;
    void init(int c){cnt=c,memset(a,0,sizeof(a));}
    void Print(){for(int i=cnt;i>=1;i--)putchar(a[i]+'0');puts("");}
    void ReadInt(int x){
        memset(buf,0,sizeof(buf));
        itoa(x,buf+1,10);
        init(strlen(buf+1));
        for(int i=1;i<=cnt;i++)a[cnt-i+1]=buf[i]-'0';
    }
};
BigData Mul(BigData A,BigData B){
    BigData C;C.init(0);
    for(int i=1;i<=A.cnt;i++)
        for(int j=1;j<=B.cnt;j++){
            C.a[i+j-1]+=A.a[i]*B.a[j];
            C.a[i+j]+=C.a[i+j-1]/10;
            C.a[i+j-1]%=10;
        }
    C.cnt=A.cnt+B.cnt-1;
    if(C.a[C.cnt+1]>=1)C.cnt++;
    while(C.a[C.cnt]==0&&C.cnt>1)C.cnt--;
    return C;
}
BigData Ans,A;
int main(){
    Ans.init(1),Ans.a[1]=1;
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)A.ReadInt(i),Ans=Mul(Ans,A);
    Ans.Print();
    return 0;
}