c语言编程问题15号之前啊~~~网上能搜到一个答案但好像不对
\u5e2e\u5fd9\u627e\u4e00\u4e0b\u9519\u8bef C\u8bed\u8a00\u7f16\u7a0b\u4f60\u628aa[N]\u6539\u6210a\u5c31\u884c\u4e86\uff0ca\u624d\u662f\u6570\u7ec4\u540d\uff0c\u4f20\u9012\u6570\u7ec4\u5143\u7d20\u9996\u5730\u5740\uff0c\u5373\u7b2c\u4e00\u4e2a\u5143\u7d20\u5730\u5740\uff0c\u4e66\u4e0a\u8fd9\u4e48\u8bf4\u7684\u3002
#include
#include
int main()
{
char a[100];
char b[100];
int d;
printf("\u8bf7\u8f93\u5165\u4f60\u8981\u8f93\u5165\u6570\u636e\u7684\u7ec4\u6570:");
scanf("%d",&d);
for (int i = 0;i < d;i++)
{
printf("\u7b2c%d\u7ec4\u6570\u636e\u4e3a:",i+1);
scanf("%s %s",a,b);
if (strcmp(a,b) > 0)
printf("-1\n");
if (strcmp(a,b) == 0)
printf("0\n");
if (strcmp(a,b) < 0)
printf("1\n");
}
return 0;
}
\u6216\u8005\u4f60\u53ef\u4ee5\u8fd9\u6837\u5199\uff0c\u8fd9\u6837\u6bd4\u8f83\u7b26\u5408acm\u7684\u5199\u6cd5\uff0c\u4e5f\u7b26\u5408\u4f60\u7684\u9898\u610f
#include
#include
int main()
{
char a[100];
char b[100];
int s;
while (scanf("%d %s %s",&s,a,b))
{
if (strcmp(a,b) > 0)
printf("-1\n");
if (strcmp(a,b) == 0)
printf("0\n");
if (strcmp(a,b) < 0)
printf("1\n");
}
return 0;
}
#include<fstream>
using namespace std;
void data(double* *a,int x,int m,int n)
{
double sum=0;
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)//两层循环控制使数组中每一个数据都成为窗口中心一次
{
switch(x)
{
case 9:
for(int y=i-1;y<=i+1;y++)
{
if(y<0||y>=m)//行超出观测区域
continue;
for(int z=j-1;z<=j+1;z++)
{
if(z<0||z>=n)//列超出观测区域
continue;
sum+=a[y][z];//sum累计窗口里所有数据的和
}
}
break;
case 25://算法同上
for(int y=i-2;y<=i+2;y++)
{
if(y<0||y>=m)
continue;
for(int z=j-2;z<=j+2;z++)
{
if(z<0||z>=n)
continue;
sum+=a[y][z];
}
}
break;
}
a[i][j]=sum/x;
sum=0;
}
}
}
int main()
{
int row,col,x;
cout<<"请输入行列数(行列数必须相同)及窗口大小："<<endl;
cin>>row>>col>>x;
ifstream in("D:/data.dat",ios::in);
ofstream out("D:/newdata.dat",ios::out);
double* *p;
p=new double* [row];//生成动态二维数组
for(int i=0;i<row;i++)
p[i]=new double[col];
for(int i=0;i<row;i++)
for(int j=0;j<col;j++)
in>>p[i][j];//读入数据
data(p,x,row,col);
for(int i=0;i<row;i++)
for(int j=0;j<col;j++)
out<<p[i][j]<<" ";//写出数据
for(int i=0;i<row;i++)
delete p[i];//释放数组所占空间
delete p;
in.close();
out.close();
}
真墨迹
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绛旓細鍑℃槸鍚︾殑鍊嶆暟鐨勪笅鏍囷紙6锛9锛12锛15锛屄仿仿凤級灏嗗叾鍏冪礌缃负0銆傚綋浠ヤ笂杩囩▼缁撴潫鍚庯紝浠嶄负1鐨勬暟缁勫厓绱犵殑涓嬫爣灏辨槸绱犳暟锛屽皢杩欎簺涓嬫爣鎵撳嵃杈撳嚭鍗冲彲銆缂栧啓绋嬪簭锛岀敤鍚湁1000涓厓绱犵殑鏁扮粍纭畾骞舵墦鍗板嚭1~999涔嬮棿鐨勬墍鏈夌礌鏁 / void Eratoshenes() { int a[1000];int i;int j;for (i=2; i<1000; i++) ...
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