9 条题解

  • 3
    @ 2021-12-15 19:27:40
    #include <iostream>
    #include <string>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    char ans[10000];
    int d = 10000;
    int n;
    int main() {
    	cin >> n;
    	memset(ans, -1, 10000);
    	ans[0] = 2;
    	for (int i = 1; i < n; i++) {
    
    		for (int j = 0; ans[j] != -1; j++) {
    			ans[j] *= 2;
    		}
    		for (int j = 0; ans[j] != -1; j++) {
    
    			if (ans[j] > 9) {
    				if (ans[j + 1] == -1) {
    					ans[j + 1] = 0;
    				}
    				ans[j + 1] += ans[j] / 10;
    				ans[j] %= 10;
    			}
    
    		}
    	}
    	int sum = 0;
    	for (int i = 0; ans[i] != -1; i++) {
    		sum++;
    	}
    	for (int i = sum - 1; i >= 0; i--) {
    		cout << (char)(ans[i] + '0');
    	}
    
    	return 0;
    }
    
    • 2
      @ 2023-11-25 20:59:40

      #include <queue> #include <math.h> #include <stack> #include <stdio.h> #include <iostream> #include <vector> #include <iomanip> #include <string.h> #include <algorithm> using namespace std; #define LL long long const int N = 1e6 + 10; const int INF = 0x3f3f3f3f; int a[N],len,x; int main() { cin >> n; a[++len]=1; while(n--) { for(int i = 1; i <= len;i++) { a[i] = a[i]*2+x; x = a[i]/10; a[i] %= 10; } if(x) a[++len]=x; x = 0; } for(int i = len;i >= 1;i--) { cout <<a[i]; }

      return 0;
      

      }

      • 1

        这道题只要用高精度算法就可以AC

        但是要注意删除前导零和进位,而且数组的第一项要设置为1

        #include<bits/stdc++.h>
        using namespace std;
        int a[100000];
        int len=2,x;
        int r,y;
        int main(){
        	cin>>r;a[1]=1;
        	while(r--){
        		for(int i=1;i<=len;i++){
        			a[i]=a[i]*2+x;
        			x=a[i]/10;
        			a[i]%=10;
        		}
        		if(x!=0)a[++len]=x;
        		x=0;
        	}while(a[len]==0){
        		len--;
        	}
        	for(int i=len;i>=1;i--)cout<<a[i];
        	return 0;
        }
        
        • 1
          @ 2023-11-25 20:59:47
          #include<iostream>
          #include<cstdio>
          #include<iomanip>
          #include<cmath>
          #include<cstring>
          #include<string>
          #include<stack>
          #include<queue>
          #define LL long long
          const int N=1e3+10;
          const int INT=0x3f3f3f3f;
          using namespace std;
          int a[N],n,len,x;
          int main(){
          	cin>>n;
          	a[++len]=1;
          	while(n--){
          		for(int i=1;i<=len;i++){
          			a[i]=a[i]*2+x;
          			x=a[i]/10;
          			a[i]%=10;
          		}
          		if(x){
          			a[++len]=x;
          		}
          		x=0;
          	}
          	for(int i=len;i>=1;i--){
          		cout<<a[i];
          	}
          	return 0;
          }
          
          • 1
            @ 2023-10-25 20:29:17

            #include <iostream> #include <string> #include <cstring> #include <algorithm> using namespace std; char ans[10000]; int d = 10000; int n; int main() { cin >> n; memset(ans, -1, 10000); ans[0] = 2; for (int i = 1; i < n; i++) {

            for (int j = 0; ans[j] != -1; j++) {
            		ans[j] *= 2;
            	}
            	for (int j = 0; ans[j] != -1; j++) {
            
            		if (ans[j] > 9) {
            			if (ans[j + 1] == -1) {
            				ans[j + 1] = 0;
            			}
            			ans[j + 1] += ans[j] / 10;
            			ans[j] %= 10;
            		}
            
            	}
            }
            int sum = 0;
            for (int i = 0; ans[i] != -1; i++) {
            	sum++;
            }
            for (int i = sum - 1; i >= 0; i--) {
            	cout << (char)(ans[i] + '0');
            }
            
            return 0;
            

            }

            • 0
              @ 2024-3-17 16:01:40
              #include <iomanip>
              #include <cmath>
              #include <cstdio>
              #include <cstring>
              #include <algorithm>
              #include <string>
              using namespace std;
              const int N=1e5+10;
              const int INF=0x3f3f3f3f;
              string a,b;
              int lena,lenb,lenc,a1[N],b1[N],c1[N],x;
              int main(){
              	cin>>a>>b;
              	lena=a.size();
              	lenb=a.size();
              	for(int i=0;i<lena;i++){
              		a1[i]=a[lena-i-1]-'0';
              	}
              	for(int i=0;i<lenb;i++){
              		b1[i]=b[lenb-i-1]-'0';
              	}
              	for(int i=0;i<lenb;i++){
              		x=0;
              		for(int j=0;j<lena;j++){
              			c1[i+j]+=a1[j]*b1[i]+x;
              			x=c1[i+j]/10;
              			c1[i+j]%=10;
              		}
              		c1[lena+i]=x;
              	}
              	lenc=lena+lenb;
              	while(c1[lenc]==0&&lenc>0){
              		lenc--;
              	}
              	for(int i=lenc;i>=0;i--){
              		cout<<c1[i];
              	}
              return 0;
              }
              
              
              • 0
                @ 2023-12-24 20:16:27
                #include<bits/stdc++.h>
                using namespace std;
                #define LL long long
                int n,a[105],len;
                int x;
                int main(){
                    cin>>n;
                    a[++len]=1;
                    while(n--){
                    	for(int i=1;i<=len;i++){
                    		a[i]=a[i]*2+x;
                    		x=a[i]/10;
                    		a[i]%=10;
                    	} 
                    	if(x){
                    		a[++len]=x;
                    	}
                    	x=0;
                    }
                    for(int i=len;i>=1;i--){
                    	cout<<a[i];
                    }
                }
                
                • 0
                  @ 2023-11-25 21:01:48
                  #include<iostream>
                  #include<cstdio>
                  #include<cmath>
                  #include<algorithm>
                  #include<string>
                  #include<cstring>
                  const int INF=0x3f3f3f3f;
                  const int N=1e5+10;
                  using namespace std;
                  int a[N],n,len,x;
                  int main(){
                  	cin>>n;
                  	a[++len]=1;
                  	while(n--)
                  	{
                  		for(int i =1;i<=len;i++ )
                  		{
                  			a[i]=a[i]*2+x;
                  			x=a[i]/10;
                  			a[i]%=10;
                  		}
                  		if(x)
                  		a[++len]=x;
                  		x=0;
                  	}
                  	for(int i=len;i>=1;i--)
                  	{
                  		cout<<a[i];
                  	}
                  	return 0;
                  }
                  
                  • 0
                    @ 2023-9-26 22:36:18
                    #include <stdio.h>
                    #include <string.h>
                    int main()
                    {
                           int n;
                           scanf("%d",&n);
                           int a[35];
                           memset(a,0,sizeof(a));
                           a[1]=1;
                           int i,j,len=1;
                           for(i=1;i<=n;i++)
                           {
                               int cf=0;
                                  for (j=1;j<=len;j++)
                                   {
                                          a[j]=a[j]*2+cf;
                                          cf=a[j]/10;
                    
                                          a[j]=a[j]%10;
                                   }
                                   while (cf!=0)
                                   {
                                          a[++len]=cf%10;
                                          cf/=10;
                                   }
                            }
                         for (i=len;i>=1;i--)
                                     printf("%d",a[i]);
                         printf("\n");
                         return 0;
                    }
                    
                    • @ 2023-9-26 22:36:39

                      当N=100时,2的N次方是一个很大的数,超出了一个长整数的表数范围。因此,为了保存2的N次方,可以定义一个数组int a[35];,每个数组元素a[i]保存结果整数的1位数,例如,保存整数1024时,a[0]=4,a[1]=2,a[2]=0,a[3]=1,并记整数的位数len=4。这样一个整数乘以2,可以将每个数组元素乘以2,同时进行进位处理即可。

                      
                      
                  • 1

                  信息

                  ID
                  1194
                  时间
                  1000ms
                  内存
                  128MiB
                  难度
                  7
                  标签
                  递交数
                  586
                  已通过
                  156
                  上传者