diff --git a/analysis table.h b/analysis table.h
new file mode 100644
index 0000000..d4d7eb4
--- /dev/null
+++ b/analysis table.h	
@@ -0,0 +1,142 @@
+#ifndef TASK1_H_INCLUDED
+#define TASK1_H_INCLUDED
+#include<stdio.h>
+#include<stdlib.h>
+void hello() {
+    printf("fsdf");
+}
+
+int mainFuntion(int in[6]) {
+    int In[6];
+    for (int i = 0; i < 6; i++) {
+        In[i] = in[i];
+    }                                                //定义输入异或
+    int k;
+    //int i;
+    void Differ(int In[6]);                           //fp是指针，用来指向FILE类型的对象。
+
+    void s_box(int array[6], int brray[4], int crray[4][16]);    //根据des把结果写成8个6比特串的并联，每一个s盒把6比特映射到4比特
+                                                             //则array是输入比特串是6位 brray是输出比特串是4位，crray是s盒4X16
+
+    printf("差分分布统计表\n");
+    printf("\n输入差分: ");
+
+    for (k = 0; k <= 5; k++) {                     //打印当前输入异或
+        printf("%d", In[k]);
+    }
+    printf("\n");
+
+    Differ(In);                       //调用差分方法
+
+    return 0;
+}
+
+
+
+void Differ(int In[6])
+{   //定义了s盒，《密码学原理与实践》P76页S1
+    int s[4][16] = { {14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7}
+                ,{0,15,7,4,14,2,13,1,10,6,12,11,9,5,3,8}
+                ,{4,1,14,8,13,6,2,11,15,12,9,7,3,10,5,0}
+                ,{15,12,8,2,4,9,1,7,5,11,3,14,10,0,6,13}
+    };
+    int b[6]; //可能的输入
+    int d[6];
+    int e[4], f[4], Out[4], n[4];//n[4]是输出差分0000~1111；
+    int flag;
+    int l;
+    int k;
+    int i, j;
+    int count;    //定义一个计数器
+    void s_box(int array[6], int brray[4], int crray[4][16]);
+
+    printf("————————————————————————————————————————————————————\n");
+    printf("S1的输出差分\t可能的输入\n");
+
+
+    for (i = 0; i <= 15; i++)
+    {
+        {
+            int t;                  //遍历输出异或
+            t = i;                    //四位二进制数0000~1111存在数组n中
+            n[3] = t % 2;  t = t / 2;
+            n[2] = t % 2;  t = t / 2;
+            n[1] = t % 2;  t = t / 2;
+            n[0] = t % 2;
+        }
+        //printf("输出差分:");
+        for (k = 0; k <= 3; k++)          //打印当前的输出异或
+        {
+            printf("%d", n[k]);
+        }
+        printf("\t\t");
+
+
+        count = 0;                   //将当前输出异或分布初始化为0次
+        for (j = 0; j <= 63; j++)              //遍历输入六元组
+        {
+
+            {   int t;
+            t = j;
+            b[5] = t % 2;  t = t / 2;
+            b[4] = t % 2;  t = t / 2;
+            b[3] = t % 2;  t = t / 2;
+            b[2] = t % 2;  t = t / 2;
+            b[1] = t % 2;  t = t / 2;
+            b[0] = t % 2;
+            }
+            for (l = 0; l <= 5; l++)
+                d[l] = b[l] ^ In[l];         //将数组进行按位异或操作，求出当前输入异或下x对应的x*，并保存在数组d[]中
+            s_box(b, e, s);               //b是输入x ，e是s盒在输入为b时的输出
+            s_box(d, f, s);               //d是b在输入异或为in[l]对应下的输入x*，f是s盒在输入为b时的输出
+            for (k = 0; k <= 3; k++)
+                Out[k] = e[k] ^ f[k];       //求出输出差分，存在Out数组中
+            flag = 1;                     //并将标志位置1；
+            for (k = 0; k <= 3; k++)
+            {
+                if (Out[k] != n[k])
+                {
+                    flag = 0;
+                    break;
+                }
+            }
+            if (flag == 1) {//符合当前输出差分的时候，当前输出差分的计数器加一
+                count = count + 1;
+                for (int z = 0; z < 6; z++) {
+                    printf("%d", b[z]);
+                }
+                printf(",");
+            }
+
+        }
+        if (count == 0)
+            printf("没有可能输入");
+        else
+            printf("\t输入数量为：%d", count);
+
+        printf("\n");
+    }
+    printf("————————————————————————————————————————————————————\n");
+}
+
+void s_box(int array[6], int brray[4], int crray[4][16])
+{
+    int p, q, t;
+    p = 2 * array[0] + array[5];                               //用b1b6比特决定s盒的行取值为p
+    q = 2 * 2 * 2 * array[1] + 2 * 2 * array[2] + 2 * array[3] + array[4];   //用b2b3b4b5决定s盒的列取值为q
+    t = crray[p][q];                                       //t就是s盒在输入为array[6]时的输出
+    brray[3] = t % 2;
+    t = t / 2;
+    brray[2] = t % 2;
+    t = t / 2;
+    brray[1] = t % 2;
+    t = t / 2;
+    brray[0] = t % 2;                                        //把t转换成二进制数存储在brray中
+}
+
+
+
+
+
+#endif // TASK1_H_INCLUDED
+#pragma once