10 条题解
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4
#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
#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; }-
周文浩 @ 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。
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ts2024stu019
LV 7
@ 2025-6-11 20:47:46
#include<bits/stdc++.h>```language
using namespace std; const int N=1e5+10; int n,a[N],yu,wei=1; int main(){ cin>>n; a[1]=1; for(int i=1;i<=n;i++){ for(int j=1;j<=wei;j++){ a[j]=a[j]*2+yu; yu=0; if(a[j]>=10){ yu=a[j]/10; a[j]%=10; } } if(yu){ wei++; a[wei]+=yu; yu=0; } } for(int i=wei;i>=1;i--){ cout<<a[i]; } return 0; } -
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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; } -
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#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; } -
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#include #include <math.h> #include #include <stdio.h> #include #include #include #include <string.h> #include 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;}
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zhangxuanju
LV 9
@ 2023-10-25 20:29:17
#include #include #include #include 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;}
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#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; } -
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#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; }
huangjiasheng
LV 9
@ 2021-12-15 19:27:40
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