C语言程序,怎么让除法输出小数得数 c语言实数除法怎样保留小数部分

C\u8bed\u8a00\u4e2d\u9664\u6cd5\u600e\u4e48\u53d6\u5f97\u5c0f\u6570

\u9664\u4e86\u4e00\u5f00\u59cb\u7528float\u8fdb\u884c\u5b9a\u4e49\u4e4b\u5916\uff0c\u540e\u9762\u8fdb\u884c\u9664\u6cd5\u8fd0\u7b97\u7684\u65f6\u5019\u8981\u52a0.0\uff0c\u5426\u5219\u7b97\u51fa\u7684\u7ed3\u679c\u7535\u8111\u4f1a\u81ea\u52a8\u53d6\u6574~~\u5982\uff1a3/2\u7684\u7ed3\u679c\u548c3.0/2\u7684\u7ed3\u679c\u5c31\u4e0d\u540c~~\u56e0\u4e3a\u6ca1\u6709\u5b9a\u4e493/2\u4e3a\u6d6e\u70b9\u578b\uff0c\u6240\u4ee53/2\u81ea\u52a8\u53d6\u6574\uff0c\u7ed3\u679c\u7b49\u4e8e1\u800c3.0/2\uff0c\u7531\u4e8e\u9884\u5148\u7528\u6d6e\u70b9\u578b\u8868\u793a\u5176\u7ed3\u679c\u663e\u7136\u4e3a\uff1a1.5 \u9664\u6cd5\u8fd0\u7b97\u7b26" / "\uff0c\u5982\u679c\u662f\u4e24\u4e2a\u6574\u6570\u76f8\u9664\u7ed3\u679c\u4e3a\u6574\u6570\u5982\u679c\u9700\u8981\u4fdd\u7559\u5c0f\u6570\u65f6 \u5fc5\u987b\u5c06\u5176\u4e2d\u4e00\u4e2a\u9664\u6570\u8f6c\u6362\u4e3a\u6d6e\u70b9\u6570 \uff03i nclude \uff03i nclude main() { float x; float y; printf("Enter x:"); scanf("%d",&x); y=fabs((5*x+1)/(x*x+1)); printf("y is %f\n",y); } \u6216\u8005 \uff03i nclude \uff03i nclude main() { int x; float y; printf("Enter x:"); scanf("%d",&x); y=fabs((float)(5*x+1)/(x*x+1)); printf("y is %f\n",y); }

#include
#include
char str[51];
int n,count,i;
int main()
{ scanf("%s",&str);
n=strlen(str);
count=1;
for(i=1;i<n;i++)
{if(str[i]!=str[i-1])
count++;
else continue; }
double ans;
ans=(float)n/count;
printf("%.2f",ans);}
printf("%.2f",ans); \u662f\u4fdd\u7559\u4e24\u4f4d\u5c0f\u6570
n\u548ccount\u90fd\u662fint\u578b\uff0c\u7528float\u5f3a\u5236\u8f6c\u6362\u6d6e\u70b9\u578b\uff0c\u624d\u80fd\u5f97\u51fa\u5e26\u5c0f\u6570\u7684\u7ed3\u679c\uff01\uff01
\u6269\u5c55\u8d44\u6599C\u8bed\u8a00\u7684\u8bbe\u8ba1\u76ee\u6807\u662f\u63d0\u4f9b\u4e00\u79cd\u80fd\u4ee5\u7b80\u6613\u7684\u65b9\u5f0f\u7f16\u8bd1\u3001\u5904\u7406\u4f4e\u7ea7\u5b58\u50a8\u5668\u3001\u4ea7\u751f\u5c11\u91cf\u7684\u673a\u5668\u7801\u4ee5\u53ca\u4e0d\u9700\u8981\u4efb\u4f55\u8fd0\u884c\u73af\u5883\u652f\u6301\u4fbf\u80fd\u8fd0\u884c\u7684\u7f16\u7a0b\u8bed\u8a00

C\u8bed\u8a00\u662f\u4e00\u95e8\u9762\u5411\u8fc7\u7a0b\u7684\u8ba1\u7b97\u673a\u7f16\u7a0b\u8bed\u8a00\uff0c\u4e0eC++\uff0cJava\u7b49\u9762\u5411\u5bf9\u8c61\u7684\u7f16\u7a0b\u8bed\u8a00\u6709\u6240\u4e0d\u540c\u3002\u5176\u7f16\u8bd1\u5668\u4e3b\u8981\u6709Clang\u3001GCC\u3001WIN-TC\u3001SUBLIME\u3001MSVC\u3001Turbo C\u7b49\u3002
C/C++\u7f16\u7a0b\u8bed\u8a00\u4e2d\uff0cint\u8868\u793a\u6574\u578b\u53d8\u91cf\uff0c\u662f\u4e00\u79cd\u6570\u636e\u7c7b\u578b\uff0c\u7528\u4e8e\u5b9a\u4e49\u4e00\u4e2a\u6574\u578b\u53d8\u91cf\uff0c\u5728\u4e0d\u540c\u7f16\u8bd1\u73af\u5883\u6709\u4e0d\u540c\u7684\u5927\u5c0f\uff0c\u4e0d\u540c\u7f16\u8bd1\u8fd0\u884c\u73af\u5883\u5927\u5c0f\u4e0d\u540c\u3002
\u8bed\u6cd5
INT\uff08number\uff09
Number \u9700\u8981\u8fdb\u884c\u5411\u4e0b\u6216\u8005\u5411\u4e0a\u820d\u5165\u53d6\u6574\u7684\u5b9e\u6570\u3002
\u8bf4\u660e
int\u51fd\u6570\u53ef\u7528floor\u51fd\u6570\u4ee3\u66ff
int(number)=floor(number,1)
\u53c2\u8003\u8d44\u6599C\u8bed\u8a00\uff3f\u767e\u5ea6\u767e\u79d1int\u51fd\u6570\uff3f\u767e\u5ea6\u767e\u79d1

ds=(float)a/(float)b;//加上类型转换就可以了。

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