TA的每日心情 | 难过 2017-6-29 01:10 |
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C语言自由的编码风格是一把双刃剑。
用不好会千人千面;用好了能结出下面这朵“奇葩”:(^_^)
大家谁感兴趣可以试试在完成相同功能的前提下用各种语言及其编码风格来PK一下。
C/C++ code
//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>
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 ;break;case 1:r =r -B ;break;case 2:r =r *B ;break;case 3:r =r /B ;break;}
switch (op2) {case 0:r =r +C ;break;case 1:r =r -C ;break;case 2:r =r *C ;break;case 3:r =r /C ;break;}
switch (op3) {case 0:r =r +D ;break;case 1:r =r -D ;break;case 2:r =r *D ;break;case 3:r =r /D ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%8d: ((%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 ;break;case 1:r1=r1-B ;break;case 2:r1=r1*B ;break;case 3:r1=r1/B ;break;}
r2=(float)C;
switch (op3) {case 0:r2=r2+D ;break;case 1:r2=r2-D ;break;case 2:r2=r2*D ;break;case 3:r2=r2/D ;break;}
switch (op2) {case 0:r =r1+r2;break;case 1:r =r1-r2;break;case 2:r =r1*r2;break;case 3:if (-0.01f<r2 && r2<0.01f) goto STEP3; r=r1/r2;break;}
if (23.99f<r && r<24.01f) {N++;printf("%8d: (%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 ;break;case 1:r =r -C ;break;case 2:r =r *C ;break;case 3:r =r /C ;break;}
switch (op1) {case 0:r =A +r ;break;case 1:r =A -r ;break;case 2:r =A *r ;break;case 3:if (-0.01f<r && r <0.01f) goto STEP4; r=A /r ;break;}
switch (op3) {case 0:r =r +D ;break;case 1:r =r -D ;break;case 2:r =r *D ;break;case 3:r =r /D ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%8d: (%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 ;break;case 1:r =r -C ;break;case 2:r =r *C ;break;case 3:r =r /C ;break;}
switch (op3) {case 0:r =r +D ;break;case 1:r =r -D ;break;case 2:r =r *D ;break;case 3:r =r /D ;break;}
switch (op1) {case 0:r =A +r ;break;case 1:r =A -r ;break;case 2:r =A *r ;break;case 3:if (-0.01f<r && r <0.01f) goto STEP5; r=A /r ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%8d: %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 ;break;case 1:r =r -D ;break;case 2:r =r *D ;break;case 3:r =r /D ;break;}
switch (op2) {case 0:r =B +r ;break;case 1:r =B -r ;break;case 2:r =B *r ;break;case 3:if (-0.01f<r && r <0.01f) continue; r=B /r ;break;}
switch (op1) {case 0:r =A +r ;break;case 1:r =A -r ;break;case 2:r =A *r ;break;case 3:if (-0.01f<r && r <0.01f) continue; r=A /r ;break;}
if (23.99f<r && r<24.01f) {N++;printf("%8d: %2d%c(%2d%c(%2d%c%2d))=24\n",N,A,opc[op1],B,opc[op2],C,opc[op3],D);}
}
}
}
}
}
}
}
}
// 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
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