diff --git a/2207/C语言同步练习源码/2.循环结构-5.c b/2207/C语言同步练习源码/2.循环结构-5.c
new file mode 100644
index 0000000..ea32b9e
--- /dev/null
+++ b/2207/C语言同步练习源码/2.循环结构-5.c
@@ -0,0 +1,31 @@
+#include <stdio.h>
+
+int main(){
+/*
+	int x, n, t=1;//tΪ2��ǰ��Ҫ�����λ�� 
+	
+	scanf("%d", &x);
+	n = x;//����֮����
+	while(t < x){//�ж�λ���Ƿ񳬹������ķ�Χ 
+		n+=t*2;
+		//���ݵ�ǰ����λ���ж�ǰһλ�Ƿ�������������һ 
+		if(n/(t*10) != x/(t*10)){
+			n-=(t*10);
+		}
+		t*=10;//����������һλ 
+	}
+	printf("%d:%d\n", x, n);
+*/
+	int x, s = 0;
+	scanf("%d", &x);
+	while(x){
+		s*=10;
+		s += (x%10+2)%10;
+		x /= 10;
+	}
+	for(;s;s/=10){
+		printf("%d", s%10);
+	}
+
+	return 0;
+}
diff --git a/2207/C语言同步练习答案.md b/2207/C语言同步练习答案.md
index 0435360..d5f5b49 100644
--- a/2207/C语言同步练习答案.md
+++ b/2207/C语言同步练习答案.md
@@ -280,6 +280,37 @@ int main(){
 ![image-20250220231006104](https://yp.smallkun.cn/markdown/image-20250220231006104.png!compress)
 
 ```c
+#include <stdio.h>
+
+int main(){
+/*
+	int x, n, t=1;//t为2当前需要加入的位置 
+	
+	scanf("%d", &x);
+	n = x;//处理之后数
+	while(t < x){//判断位置是否超过的数的范围 
+		n+=t*2;
+		//根据当前输入位置判断前一位是否溢出如果溢出则减一 
+		if(n/(t*10) != x/(t*10)){
+			n-=(t*10);
+		}
+		t*=10;//继续处理下一位 
+	}
+	printf("%d:%d\n", x, n);
+*/
+	int x, s = 0;
+	scanf("%d", &x);
+	while(x){
+		s*=10;
+		s += (x%10+2)%10;
+		x /= 10;
+	}
+	for(;s;s/=10){
+		printf("%d", s%10);
+	}
+
+	return 0;
+}
 ```
 
 ### 循环结构-6