c语言怎样精确到小数点后8位. C语言输入小数,精确到小数点后10位

c\u8bed\u8a00\u4e2d\u5982\u4f55\u5224\u65ad\u8ba1\u7b97\u7ed3\u679c\u7cbe\u786e\u5230\u5c0f\u6570\u70b9\u540e8\u4f4d

int multiply(int i)
{
if (i ==1 || i == 0) return 1;
else return (i*multiply(i-1));

}
int _tmain(int argc, _TCHAR* argv[])
{
int i=1,j=1;
double x;
scanf("%lf",&x);
double result=1;

while (1)
{
j = j*(-1);
int sum = multiply(i);
double k;
k=pow(x,2*i)/sum;
result =result+j*pow(x,2*i)/(double)sum;
if (k<=0.00000001)
{printf("result = %.8f ; i = %d",result, i);
break;}

i++;
}
system("pause");
return 0;
}

\u628ascanf("%15.10f",&x);\u5e94\u8be5\u6539\u6210scanf("%15lf",&x);\u8fd9\u6837\u5c31\u597d\u4e86\u3002
\u4ee3\u7801\u4e2dx\u662fdouble\u578b\u7684\uff0c\u5728\u8f93\u5165\u6570\u636e\u7684\u65f6\u5019\u662f\u4e0d\u5141\u8bb8\u4f7f\u7528%f\u7684\uff0cdouble\u578b\u7684\u6570\u636e\u8f93\u5165\u5fc5\u987b\u4f7f\u7528%lf
\u8fd8\u6709\uff0cscanf("%15.10f",&x);15.10\u8fd9\u4e2a\u662f\u4e0d\u5141\u8bb8\u51fa\u73b0\u7684\uff0c\u5728\u6570\u636e\u8f93\u5165\u7684\u65f6\u5019\u53ea\u80fd\u6307\u5b9a\u8f93\u5165\u6570\u636e\u7684\u603b\u957f\u5ea6\uff0c\u4e0d\u80fd\u6307\u5b9a\u5c0f\u6570\u70b9\u540e\u7684\u957f\u5ea6\u3002

\u6269\u5c55\u8d44\u6599\uff1a
scanf(\u51fd\u6570\u539f\u578b
int scanf(const char * restrict format,...);
\u51fd\u6570 scanf() \u662f\u4ece\u6807\u51c6\u8f93\u5165\u6d41stdin(\u6807\u51c6\u8f93\u5165\u8bbe\u5907\uff0c\u4e00\u822c\u6307\u5411\u952e\u76d8)\u4e2d\u8bfb\u5185\u5bb9\u7684\u901a\u7528\u5b50\u7a0b\u5e8f\uff0c\u53ef\u4ee5\u8bf4\u660e\u7684\u683c\u5f0f\u8bfb\u5165\u591a\u4e2a\u5b57\u7b26\uff0c\u5e76\u4fdd\u5b58\u5728\u5bf9\u5e94\u5730\u5740\u7684\u53d8\u91cf\u4e2d\u3002
\u51fd\u6570\u7684\u7b2c\u4e00\u4e2a\u53c2\u6570\u662f\u683c\u5f0f\u5b57\u7b26\u4e32\uff0c\u5b83\u6307\u5b9a\u4e86\u8f93\u5165\u7684\u683c\u5f0f\uff0c\u5e76\u6309\u7167\u683c\u5f0f\u8bf4\u660e\u7b26\u89e3\u6790\u8f93\u5165\u5bf9\u5e94\u4f4d\u7f6e\u7684\u4fe1\u606f\u5e76\u5b58\u50a8\u4e8e\u53ef\u53d8\u53c2\u6570\u5217\u8868\u4e2d\u5bf9\u5e94\u7684\u6307\u9488\u6240\u6307\u4f4d\u7f6e\u3002\u6bcf\u4e00\u4e2a\u6307\u9488\u8981\u6c42\u975e\u7a7a\uff0c\u5e76\u4e14\u4e0e\u5b57\u7b26\u4e32\u4e2d\u7684\u683c\u5f0f\u7b26\u4e00\u4e00\u987a\u6b21\u5bf9\u5e94\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-scanf \uff08\u8ba1\u7b97\u673a\u8bed\u8a00\u51fd\u6570\uff09

s 和i是int类型的,3/2 是=1的,如果是float 或者double 3.0/2.0=1.5的

这么改t=(double)s/i;,先把除数变为double

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