求 SN = 2 + 22 + 222 + 2222 + ... + 2...2（n个2）的程序

C语言求 SN = 2 + 22 + 222 + 2222 + … + 2…2（n个2）的程序

一、基础方法（逐项累加）

1. 非函数方式

#include<stdio.h>intmain(){intn,i,j;longlongsum=0,term=0;printf("请输入n的值: ");scanf("%d",&n);for(i=1;i<=n;i++){term=0;for(j=1;j<=i;j++){term=term*10+2;}sum+=term;}printf("SN = %lld\n",sum);return0;}

2. 函数方式

#include<stdio.h>longlongcalculateSN(intn){longlongsum=0,term=0;inti,j;for(inti=1;i<=n;i++){term=0;for(intj=1;j<=i;j++){term=term*10+2;}sum+=term;}returnsum;}intmain(){intn;printf("请输入n的值: ");scanf("%d",&n);printf("SN = %lld\n",calculateSN(n));return0;}

二、优化方法（递推公式）

1. 非函数方式

#include<stdio.h>intmain(){intn,i;longlongsum=0,term=0;printf("请输入n的值: ");scanf("%d",&n);for(i=1;i<=n;i++){term=term*10+2;sum+=term;}printf("SN = %lld\n",sum);return0;}

2. 函数方式

#include<stdio.h>longlongcalculateSN(intn){longlongsum=0,term=0;for(inti=1;i<=n;i++){term=term*10+2;sum+=term;}returnsum;}intmain(){intn;printf("请输入n的值: ");scanf("%d",&n);printf("SN = %lld\n",calculateSN(n));return0;}

三、递归方法

1. 递归求和

#include<stdio.h>longlonggetTerm(intn){if(n==1)return2;returngetTerm(n-1)*10+2;}longlongcalculateSN(intn){if(n==1)return2;returncalculateSN(n-1)+getTerm(n);}intmain(){intn;printf("请输入n的值: ");scanf("%d",&n);printf("SN = %lld\n",calculateSN(n));return0;}

2. 递归优化版

#include<stdio.h>voidcalculateSN(intn,longlong*sum,longlong*term){if(n==0)return;*term=*term*10+2;*sum+=*term;calculateSN(n-1,sum,term);}intmain(){intn;longlongsum=0,term=0;printf("请输入n的值: ");scanf("%d",&n);calculateSN(n,&sum,&term);printf("SN = %lld\n",sum);return0;}

四、数学公式方法

1. 等比数列公式

#include<stdio.h>#include<math.h>longlongcalculateSN(intn){inti;// SN = 2/9 * (10^n - 1) + 20/9 * (10^(n-1) - 1) + ... + 2n/9 * (10 - 1)// 简化后：SN = 2/81 * (10^(n+1) - 9n - 10)longlongpower=1;for(i=0;i<=n+1;i++){power*=10;}return(2*(power-9*n-10))/81;}intmain(){intn;printf("请输入n的值: ");scanf("%d",&n);printf("SN = %lld\n",calculateSN(n));return0;}

2. 递推公式优化

#include<stdio.h>longlongcalculateSN(intn){inti;longlongsum=0,term=0;for(i=1;i<=n;i++){term=term*10+2;sum+=term;}returnsum;}intmain(){intn,i;printf("请输入n的值: ");scanf("%d",&n);printf("各项为: ");longlongterm=0;for(i=1;i<=n;i++){term=term*10+2;printf("%lld ",term);}printf("\n");printf("SN = %lld\n",calculateSN(n));return0;}

五、不同实现风格

1. 数组存储方式

#include<stdio.h>intmain(){intn,i;longlongterms[100],sum=0;printf("请输入n的值: ");scanf("%d",&n);terms[0]=2;sum=terms[0];for(i=1;i<n;i++){terms[i]=terms[i-1]*10+2;sum+=terms[i];}printf("各项为: ");for(i=0;i<n;i++){printf("%lld ",terms[i]);}printf("\n");printf("SN = %lld\n",sum);return0;}

2. 指针方式

#include<stdio.h>voidcalculateSN(intn,longlong*sum){inti;longlongterm=0;for(i=1;i<=n;i++){term=term*10+2;*sum+=term;}}intmain(){intn;longlongsum=0;printf("请输入n的值: ");scanf("%d",&n);calculateSN(n,&sum);printf("SN = %lld\n",sum);return0;}

3. 宏定义方式

#include<stdio.h>#defineCALCULATE_SN(n)({\inti;longlongsum=0,term=0;\for(i=1;i<=(n);i++){\ term=term*10+2;\ sum+=term;\}\ sum;\})intmain(){intn;printf("请输入n的值: ");scanf("%d",&n);printf("SN = %lld\n",CALCULATE_SN(n));return0;}

六、完整测试程序（多种算法对比）

#include<stdio.h>// 方法1：双重循环longlongmethod1(intn){longlongsum=0,term=0;inti,j;for(i=1;i<=n;i++){term=0;for(j=1;j<=i;j++){term=term*10+2;}sum+=term;}returnsum;}// 方法2：递推公式longlongmethod2(intn){longlongsum=0,term=0;inti;for(i=1;i<=n;i++){term=term*10+2;sum+=term;}returnsum;}// 方法3：递归longlongmethod3(intn){if(n==0)return0;longlongterm=0;inti;for(i=1;i<=n;i++){term=term*10+2;}returnterm+method3(n-1);}intmain(){intn,i;printf("请输入n的值: ");scanf("%d",&n);printf("\n=== 不同算法结果 ===\n");printf("双重循环: SN = %lld\n",method1(n));printf("递推公式: SN = %lld\n",method2(n));printf("递归方法: SN = %lld\n",method3(n));// 显示各项printf("\n各项为: ");longlongterm=0;for(i=1;i<=n;i++){term=term*10+2;printf("%lld ",term);}printf("\n");return0;}

输入输出示例

输入：

请输入n的值: 3

输出：

SN = 246

解释：SN = 2 + 22 + 222 = 246

输入：

请输入n的值: 5

输出：

SN = 24690

解释：SN = 2 + 22 + 222 + 2222 + 22222 = 24690

算法复杂度对比

推荐使用

递推公式方法是最推荐的实现方式：

• 时间复杂度：O(n)
• 空间复杂度：O(1)
• 代码简洁，易于理解
• 性能优秀