C++-蓝桥杯-入门训练

 1 #include <cstdio>
 2 struct Matrix{int a[5][5];};
 3 const int N=2,MOD=10007;
 4 Matrix A,B,O,I;
 5 Matrix Mul(Matrix A,Matrix B){
 6     Matrix C=O;
 7     for(int i=1;i<=N;i++)
 8         for(int j=1;j<=N;j++)
 9             for(int k=1;k<=N;k++)
10                 C.a[i][j]=(C.a[i][j]+A.a[i][k]*B.a[k][j])%MOD;
11     return C;
12 }
13 Matrix Pow(Matrix A,int n){
14     Matrix B=I;
15     for(;n;n>>=1,A=Mul(A,A))if(n&1)B=Mul(B,A);
16     return B;
17 }
18 int main(){
19     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);
20     A.a[1][1]=0,A.a[1][2]=1;
21     A.a[2][1]=1,A.a[2][2]=1;
22     int n;scanf("%d",&n),B=Pow(A,n+1),printf("%d\n",B.a[1][1]); 
23     return 0;
24 }

1 #include <cmath>
2 #include <cstdio>
3 typedef double db;
4 db r,pi=acos(-1);
5 int main(){
6     scanf("%lf",&r),printf("%.7lf",pi*r*r);
7     return 0;
8 }

1 #include <cstdio>
2 typedef long long ll;
3 ll ans,n;
4 int main(){
5     scanf("%lld",&n),printf("%lld",(1+n)*n/2);
6     return 0;
7 }

1 #include <cstdio>
2 int main(){
3     int a,b;scanf("%d%d",&a,&b),printf("%d",a+b);
4     return 0;
5 }

 1 #include <queue>
 2 #include <cstdio>
 3 using namespace std;
 4 priority_queue<int>q;
 5 int main(){
 6     int n,a;scanf("%d",&n);
 7     for(int i=1;i<=n;i++)scanf("%d",&a),q.push(-a);
 8     for(int i=1;i<=n;i++)printf("%d ",-q.top()),q.pop();
 9     return 0;
10 }

 1 #include <cstdio>
 2 #include <string>
 3 #include <cstring>
 4 #include <iostream>
 5 using namespace std;
 6 char s[100010];
 7 int main(){
 8     int n,len,num,ok;scanf("%d",&n);
 9     for(int i=1;i<=n;i++){
10         scanf("%s",s+1),len=strlen(s+1);
11         string ans="";
12         for(int i=1;i<=len;i++)
13             switch(s[i]){
14                 case ‘0‘:ans+="0000";break;case ‘1‘:ans+="0001";break;
15                 case ‘2‘:ans+="0010";break;case ‘3‘:ans+="0011";break;
16                 case ‘4‘:ans+="0100";break;case ‘5‘:ans+="0101";break;
17                 case ‘6‘:ans+="0110";break;case ‘7‘:ans+="0111";break;
18                 case ‘8‘:ans+="1000";break;case ‘9‘:ans+="1001";break;
19                 case ‘A‘:ans+="1010";break;case ‘B‘:ans+="1011";break;
20                 case ‘C‘:ans+="1100";break;case ‘D‘:ans+="1101";break;
21                 case ‘E‘:ans+="1110";break;case ‘F‘:ans+="1111";break;
22             }//printf("%s\n",ans.c_str());            
23         switch(len%3){
24             case 1:ans="00"+ans;break;
25             case 2:ans="0"+ans;break;
26         }
27         len=ans.length();ok=0;
28         for(int i=0;i<len;i+=3){
29             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]);
30             if(num)ok=1;if(ok)putchar(num+‘0‘);
31         }
32         puts("");
33     }
34     return 0;
35 }

 1 #include <cstdio>
 2 #include <string>
 3 #include <cstring>
 4 #include <cstdlib>
 5 using namespace std;
 6 const int MAXN=5000;
 7 char buf[MAXN];
 8 struct BigData{
 9     int a[MAXN],cnt;
10     void init(int c){cnt=c,memset(a,0,sizeof(a));}
11     void Print(){for(int i=cnt;i>=1;i--)putchar(a[i]+‘0‘);puts("");}
12     void ReadInt(int x){
13         memset(buf,0,sizeof(buf));
14         itoa(x,buf+1,10);
15         init(strlen(buf+1));
16         for(int i=1;i<=cnt;i++)a[cnt-i+1]=buf[i]-‘0‘;
17     }
18 };
19 BigData Mul(BigData A,BigData B){
20     BigData C;C.init(0);
21     for(int i=1;i<=A.cnt;i++)
22         for(int j=1;j<=B.cnt;j++){
23             C.a[i+j-1]+=A.a[i]*B.a[j];
24             C.a[i+j]+=C.a[i+j-1]/10;
25             C.a[i+j-1]%=10;
26         }
27     C.cnt=A.cnt+B.cnt-1;
28     if(C.a[C.cnt+1]>=1)C.cnt++;
29     while(C.a[C.cnt]==0&&C.cnt>1)C.cnt--;
30     return C;
31 }
32 BigData Ans,A;
33 int main(){
34     Ans.init(1),Ans.a[1]=1;
35     int n;scanf("%d",&n);
36     for(int i=1;i<=n;i++)A.ReadInt(i),Ans=Mul(Ans,A);
37     Ans.Print();
38     return 0;
39 }