优化代码:int a1[]={ b2[0], b2[1], b2[2], b2[3], b1[0], b1[1], b1[2], b1[3], b2[4], b2[5], b2[6], b2[7], b1[4], b1[5], b1[6], b1[7], b2[8], b2[9], b2[10], b2[11], b1[8], b1[9], b1[10], b1[11], b2[12], b2[13], b2[14], b2[15], b1[12], b1[13], b1[14], b1[15], b2[16], b2[17], b2[18], b2[19], b1[16], b1[17], b1[18], b1[19], b2[20], b2[21], b2[22], b2[23], b1[20], b1[21], b1[22], b1[23], b2[24], b2[25], b2[26], b2[27], b1[24], b1[25], b1[26], b1[27], b2[28], b2[29], b2[30], b2[31], b1[28], b1[29], b1[30], b1[31], };
时间: 2023-08-17 13:16:15 浏览: 31
可以考虑使用循环来简化代码,如下所示:
```
int a1[64];
int index = 0;
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 4; j++) {
a1[index++] = b2[i*4+j];
}
for (int j = 0; j < 4; j++) {
a1[index++] = b1[i*4+j];
}
}
```
这样可以将代码简化,并且使得代码可读性更好。同时,如果需要修改数组长度,只需要修改循环的参数即可,不需要手动修改每个元素的下标。
相关问题
优化if((a1=b1)&&(a2=b2)&&(a3=b3))
如果您的问题是如何优化代码 `if((a1=b1) && (a2=b2) && (a3=b3))`,那么可以考虑以下优化:
1. 将赋值操作和判断操作分开,这样可以增加代码可读性和可维护性:
```if(a1==b1 && a2==b2 && a3==b3)```
2. 如果需要同时对多个变量进行判断,可以使用循环来遍历变量数组:
```
bool isEqual = true;
for(int i=0; i<3; i++){
if(a[i] != b[i]){
isEqual = false;
break;
}
}
if(isEqual){
// do something
}
```
3. 如果只需要判断两个数组是否相等,可以使用 `std::equal` 函数:
```
if(std::equal(a, a+3, b)){
// do something
}
```
帮我生成这个代码的流程图:a1,b1,c1=input().split(" ") a2,b2,c2=input().split(" ") a1=int(a1) b1=int(b1) c1=int(c1) a2=int(a2) b2=int(b2) c2=int(c2) d1=[0,31,28,31,30,31,30,31,31,30,31,30,31] d2=[0,31,29,31,30,31,30,31,31,30,31,30,31] sum=0 def leapyear(x): if x%400==0 or (x%4==0 and x%100!=0): return True else: return False def cha(x,x1,x2): cum=0 if leapyear(x): for i in range(x1,x2): cum+=d2[i] else: for i in range(x1,x2): cum+=d1[i] return cum if a1==a2: if b1==b2: if c1==c2: sum+=0 else: sum+=(c2-c1) else: sum+=(cha(a1,b1,b2)+c2-c1) else: sum+=(cha(a1,b1,12)+31-c1+cha(a2,1,b2)+c2-1+1) week=sum//7+1 weekdays=sum%7+1 print(f"{week} {weekdays}")
好的,以下是代码流程图:
```
+------------------------+
| Start |
+------------------------+
|
+------------------------+
| Input a1,b1,c1 |
+------------------------+
|
+------------------------+
| Input a2,b2,c2 |
+------------------------+
|
+------------------------+
| Convert a1,b1,c1 to int |
+------------------------+
|
+------------------------+
| Convert a2,b2,c2 to int |
+------------------------+
|
+------------------------+
| Initialize d1,d2,sum |
+------------------------+
|
+------------------------+
+------------>| Define leapyear function |
| +------------------------+
| |
| +------------------------+
| | Define cha function |
| +------------------------+
| |
+-------------+---------------------------+
| If a1==a2 |
| If b1==b2 |
| If c1==c2 |
| sum+=0 |
| Else |
| sum+=(c2-c1) |
| Else |
| sum+=(cha(a1,b1,b2)+c2-c1) |
| Else |
| sum+=(cha(a1,b1,12)+31-c1+cha(a2,1,b2)+c2-1+1) |
+------------------------+
|
+------------------------+
| Calculate week |
+------------------------+
|
+------------------------+
| Calculate weekdays |
+------------------------+
|
+------------------------+
| Output week and weekdays |
+------------------------+
|
+------------------------+
| End |
+------------------------+
```
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