即：假设输入的值为10，则：
10 10 10 10 = 1
10 10 10 10 = 2
10 10 10 10 = 3
10 10 10 10 = 4
10 10 10 10 = 5
10 10 10 10 = 6
10 10 10 10 = 7
10 10 10 10 = 8
10 10 10 10 = 9
10 10 10 10 = 10
只要每一个式子给出一个式子即可。若不存在，则输出该式子无法给成。
你这个思路适合用递归函数实现，其实就是对一个数进行自身计算
仅供参考：
//定义下面5种运算顺序：
// ((A @ B) @ C) @ D =
// (A @ B) @ (C @ D) =
// (A @ (B @ C)) @ D =
// A @ ((B @ C) @ D) =
// A @ (B @ (C @ D))=
//23.99<计算结果<24.01
//其中：
//1≤A、B、C、D≤13
//@为“+、-、*、/”之一
//用浮点数计算，循环遍历以上所有情况即可。
#include <stdio.h>
#define when break;case
int A,B,C,D,N;
float r,r1,r2;
char op1,op2,op3;
char opc[4]={'+','-','*','/'};
void main() {
N=0;
for (A =1;A <=13;A ++) {
for (B =1;B <=13;B ++) {
for (C =1;C <=13;C ++) {
for (D =1;D <=13;D ++) {
for (op1=0;op1< 4;op1++) {
for (op2=0;op2< 4;op2++) {
for (op3=0;op3< 4;op3++) {
// ((A @ B) @ C) @ D
r=(float)A;
switch (op1) {case 0:r =r +B ;when 1:r =r -B ;when 2:r =r *B ;when 3: r =r /B ;break;}
switch (op2) {case 0:r =r +C ;when 1:r =r -C ;when 2:r =r *C ;when 3: r =r /C ;break;}
switch (op3) {case 0:r =r +D ;when 1:r =r -D ;when 2:r =r *D ;when 3: r =r /D ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%5d: ((%2d %c %2d) %c %2d) %c %2d =24\n",N,A,opc[op1],B,opc[op2],C,opc[op3],D);}
// (A @ B) @ (C @ D)
r1=(float)A;
switch (op1) {case 0:r1=r1+B ;when 1:r1=r1-B ;when 2:r1=r1*B ;when 3: r1=r1/B ;break;}
r2=(float)C;
switch (op3) {case 0:r2=r2+D ;when 1:r2=r2-D ;when 2:r2=r2*D ;when 3: r2=r2/D ;break;}
switch (op2) {case 0:r =r1+r2;when 1:r =r1-r2;when 2:r =r1*r2;when 3:if (-0.01f<r2 && r2<0.01f) goto STEP3;r =r1/r2;break;}
if (23.99f<r && r<24.01f) {N++;printf("%5d: (%2d %c %2d) %c (%2d %c %2d) =24\n",N,A,opc[op1],B,opc[op2],C,opc[op3],D);}
// (A @ (B @ C)) @ D
STEP3:
r=(float)B;
switch (op2) {case 0:r =r +C ;when 1:r =r -C ;when 2:r =r *C ;when 3: r =r /C ;break;}
switch (op1) {case 0:r =A +r ;when 1:r =A -r ;when 2:r =A *r ;when 3:if (-0.01f<r && r <0.01f) goto STEP4;r =A /r ;break;}
switch (op3) {case 0:r =r +D ;when 1:r =r -D ;when 2:r =r *D ;when 3: r =r /D ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%5d: (%2d %c (%2d %c %2d)) %c %2d =24\n",N,A,opc[op1],B,opc[op2],C,opc[op3],D);}
// A @ ((B @ C) @ D)
STEP4:
r=(float)B;
switch (op2) {case 0:r =r +C ;when 1:r =r -C ;when 2:r =r *C ;when 3: r =r /C ;break;}
switch (op3) {case 0:r =r +D ;when 1:r =r -D ;when 2:r =r *D ;when 3: r =r /D ;break;}
switch (op1) {case 0:r =A +r ;when 1:r =A -r ;when 2:r =A *r ;when 3:if (-0.01f<r && r <0.01f) goto STEP5;r =A /r ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%5d: %2d %c ((%2d %c %2d) %c %2d) =24\n",N,A,opc[op1],B,opc[op2],C,opc[op3],D);}
// A @ (B @ (C @ D))
STEP5:
r=(float)C;
switch (op3) {case 0:r =r +D ;when 1:r =r -D ;when 2:r =r *D ;when 3: r =r /D ;break;}
switch (op2) {case 0:r =B +r ;when 1:r =B -r ;when 2:r =B *r ;when 3:if (-0.01f<r && r <0.01f) continue ;r =B /r ;break;}
switch (op1) {case 0:r =A +r ;when 1:r =A -r ;when 2:r =A *r ;when 3:if (-0.01f<r && r <0.01f) continue ;r =A /r ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%5d: %2d %c (%2d %c (%2d %c %2d))=24\n",N,A,opc[op1],B,opc[op2],C,opc[op3],D);}
}//op3
}//op2
}//op1
}//D
}//C
}//B
}//A
}
//运行结果：
// 1: (( 1 + 1) + 1) * 8 =24
// 2: ( 1 + ( 1 + 1)) * 8 =24
// 3: ( 1 + 1) * ( 1 + 11) =24
// ... ...
//53280: (13 + 13) / (13 / 12) =24
//53281: (13 - (13 / 13)) + 12 =24
//53282: 13 - ((13 / 13) - 12) =24