c语言编程题:编写函数squaresum求解两个整数的平方和，并在main当中输入两个整数验证结果 c语言编程中，sinx怎么表示？
c\u8bed\u8a00\u7f16\u7a0b \u628a\u5341\u8fdb\u5236\u8f6c\u6362\u4e3a\u5341\u516d\u8fdb\u5236\u65b9\u6cd5\u4e00\uff1a#include#include using namespace std;
string DecIntToHexStr(long long num){string str;long long Temp = num / 16;int left = num % 16;if (Temp > 0)str +=
DecIntToHexStr(Temp);if (left < 10)str += (left + '0');elsestr += ('A' + left - 10);return str;}string DecStrToHexStr(string str){long long Dec = 0;for (int i = 0; i < str.size(); ++i)Dec =
Dec * 10 + str[i] - '0';return DecIntToHexStr(Dec);}int main(){string DecStr;while (cin >> DecStr){cout << "0x" + DecStrToHexStr(DecStr);}return 0;}
\u65b9\u6cd5\u4e8c\uff1a
#include#include #include
#include #include using namespace std;int main()
{string decStr;cin >> decStr;int num = stoi(decStr);vectorv;while (num){int bit = num % 2;v.push_back(bit);num = num / 2;}
reverse(v.begin(), v.end());string hexStr;if (v.size() % 4 == 0){for (int i = 0; i < v.size() / 4; ++i){int temp = 0, k = 3;for (int j = i * 4; j < (i +
1) * 4; ++j){if (v[j] == 1)temp += pow(2, k);--k;} if (temp <= 9)hexStr += temp;elsehexStr += ('A' + temp - 10);}}
else{int zeroNum = 4 - v.size() % 4;for (int i = 0; i < zeroNum; ++i)v.insert(v.begin(), 0);for (int i = 0; i < v.size() / 4; ++i){int temp = 0, k = 3;for
(int j = i * 4; j < (i + 1) * 4; ++j){if (v[j] == 1)temp += pow(2, k);--k;}if (temp <= 9){stringstream ss;string strTemp;
ss > strTemp;hexStr += strTemp;}elsehexStr += ('A' + temp - 10);}}cout << hexStr << endl;return 0;}
\u6269\u5c55\u8d44\u6599\uff1a
\u8fd8\u67093\u79cd\u65b9\u5f0f\u53ef\u4ee5\u5b9e\u73b0\uff0c\u5176\u4e2d\u4e24\u79cd\u662f\u4f7f\u7528\u7cfb\u7edf\u51fd\u6570\uff0c\u53e6\u4e00\u79cd\u662f\u76f4\u63a5\u81ea\u5df1\u7f16\u5199\u3002
\u4e0b\u9762\u7684\u4ee3\u7801\u5c31\u662f3\u79cd\u65b9\u5f0f\u7684\u5b9e\u73b0\uff0c\u5305\u62ec2\u4f4d\u7684\u6574\u6570\u548c\u4efb\u610f\u6574\u6570\uff082\u768431\u6b21\u65b9\u4ee5\u5185\uff09\u90fd\u53ef\u4ee5\u3002\u53ef\u81ea\u5df1\u9009\u62e9\u9700\u8981\u7684\u5b9e\u73b0\u65b9\u5f0f\u3002
\u5229\u7528\u7cfb\u7edf\u51fd\u6570\u6709
1. char * itoa ( int value, char * str, int base );value\u662f\u8981\u8f6c\u5316\u7684\u6570\u5b57\uff0cstr\u662f\u8f6c\u5316\u540e\u7684\u5b57\u7b26\u4e32\u5b58\u50a8\u7684\u4f4d\u7f6e\uff0cbase\u662f\u8fdb\u5236\u6570\uff08\u4f46\u662f\u8fd9\u4e2a\u51fd\u6570\u4e0d\u662f\u6807\u51c6C\u51fd\u6570\uff0c\u6709\u4e9b\u7f16\u8bd1\u5668\u662f\u4e0d\u652f\u6301\u7684\uff01\uff09\u3002\u6240\u4ee5\u4ee3\u7801\u53ef\u4ee5\u4e3a\uff1a
char buffer [33]; //\u7528\u4e8e\u5b58\u653e\u8f6c\u6362\u597d\u7684\u5341\u516d\u8fdb\u5236\u5b57\u7b26\u4e32\uff0c\u53ef\u6839\u636e\u9700\u8981\u5b9a\u4e49\u957f\u5ea6 char * inttohex(int aa){ itoa (aa, buffer, 16); return (buffer);}
2. sprintf(str,"%x",value);str\u662f\u8f6c\u5316\u5b58\u50a8\u7684\u4f4d\u7f6e\uff0c%x\u8868\u793a\u5341\u516d\u8fdb\u5236\u683c\u5f0f\uff0cvalue\u662f\u8981\u8f6c\u5316\u7684\u6570\u5b57\u3002\u6240\u4ee5\u4ee3\u7801\u53ef\u4ee5\u4e3a\uff1a
char buffer [33]; //\u7528\u4e8e\u5b58\u653e\u8f6c\u6362\u597d\u7684\u5341\u516d\u8fdb\u5236\u5b57\u7b26\u4e32\uff0c\u53ef\u6839\u636e\u9700\u8981\u5b9a\u4e49\u957f\u5ea6 char * inttohex(int aa){ sprintf(buffer, "%x", aa); return (buffer);}
3. \u81ea\u5df1\u7f16\u5199
\u5982\u679c\u53c2\u6570\u53ea\u8981\u4e24\u4f4d\u6570\u7684\u6574\u6570\uff0c\u90a3\u4e48\u5f88\u7b80\u5355\u3002
\u4ee3\u7801\u5982\u4e0b\uff1a
#include #include char buffer [33]; //\u7528\u4e8e\u5b58\u653e\u8f6c\u6362\u597d\u7684\u5341\u516d\u8fdb\u5236\u5b57\u7b26\u4e32\uff0c\u53ef\u6839\u636e\u9700\u8981\u5b9a\u4e49\u957f\u5ea6 char * inttohex(int aa)
{ sprintf(buffer, "%x", aa); return (buffer);} int main ()
{ int num; char * hex_str; printf ("Enter a number: "); scanf ("%d",&num); hex_str = inttohex (num); printf ("Hexadecimal number: %sH\n", hex_str); return 0;}
\u5728\u5199C\u8bed\u8a00\u7684\u7a0b\u5e8f\u65f6\uff0c\u5728\u5f00\u5934\u52a0\u4e0a\u4e00\u4e2a\u5934\u6587\u4ef6math.h\u5373\u53ef\u3002
\u5373\u53ef\u76f4\u63a5\u4f7f\u7528sin(x),\u7279\u522b\u6ce8\u610fx\u5e94\u8be5\u4e3a\u5f27\u5ea6\u5236\uff0c\u5982\u679c\u4e0d\u662f\u5f27\u5ea6\u5236\u9700\u8981\u8f6c\u5316\u4e3a\u5f27\u5ea6\u5236\u3002
\u6dfb\u52a0\u5934\u6587\u4ef6\u65b9\u6cd5\uff1a#include\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u5728C\u8bed\u8a00\u5bb6\u65cf\u7a0b\u5e8f\u4e2d\uff0c\u5934\u6587\u4ef6\u88ab\u5927\u91cf\u4f7f\u7528\u3002\u4e00\u822c\u800c\u8a00\uff0c\u6bcf\u4e2aC++/C\u7a0b\u5e8f\u901a\u5e38\u7531\u5934\u6587\u4ef6\u548c\u5b9a\u4e49\u6587\u4ef6\u7ec4\u6210\u3002\u5934\u6587\u4ef6\u4f5c\u4e3a\u4e00\u79cd\u5305\u542b\u529f\u80fd\u51fd\u6570\u3001\u6570\u636e\u63a5\u53e3\u58f0\u660e\u7684\u8f7d\u4f53\u6587\u4ef6\uff0c\u4e3b\u8981\u7528\u4e8e\u4fdd\u5b58\u7a0b\u5e8f\u7684\u58f0\u660e\uff0c\u800c\u5b9a\u4e49\u6587\u4ef6\u7528\u4e8e\u4fdd\u5b58\u7a0b\u5e8f\u7684\u5b9e\u73b0\u3002
C\u6807\u51c6\u51fd\u6570\u5e93\uff08C Standard library\uff09\u662f\u6240\u6709\u7b26\u5408\u6807\u51c6\u7684\u5934\u6587\u4ef6\uff08head file\uff09\u7684\u96c6\u5408\uff0c\u4ee5\u53ca\u5e38\u7528\u7684\u51fd\u6570\u5e93\u5b9e\u73b0\u7a0b\u5e8f\uff0c\u4f8b\u5982I/O \u8f93\u5165\u8f93\u51fa\u548c\u5b57\u7b26\u4e32\u63a7\u5236\u3002
\u4e0d\u50cf COBOL\u3001Fortran \u548c PL/I\u7b49\u7f16\u7a0b\u8bed\u8a00\uff0c\u5728 C \u8bed\u8a00\u7684\u5de5\u4f5c\u4efb\u52a1\u91cc\u4e0d\u4f1a\u5305\u542b\u5d4c\u5165\u7684\u5173\u952e\u5b57\uff0c\u6240\u4ee5\u51e0\u4e4e\u6240\u6709\u7684 C \u8bed\u8a00\u7a0b\u5e8f\u90fd\u662f\u7531\u6807\u51c6\u51fd\u6570\u5e93\u7684\u51fd\u6570\u6765\u521b\u5efa\u7684\u3002
1995\u5e74\uff0cNormative Addendum 1 (NA1)\u6279\u51c6\u4e86\u4e09\u4e2a\u5934\u6587\u4ef6(iso646.h, wchar.h, and wctype.h)\u589e\u52a0\u5230C\u6807\u51c6\u51fd\u6570\u5e93\u4e2d\u3002C99\u6807\u51c6\u589e\u52a0\u4e86\u516d\u4e2a\u5934\u6587\u4ef6(complex.h, fenv.h, inttypes.h, stdbool.h, stdint.h, and tgmath.h)\u3002
C11\u6807\u51c6\u4e2d\u53c8\u65b0\u589e\u4e865\u4e2a\u5934\u6587\u4ef6(stdalign.h, stdatomic.h, stdnoreturn.h, threads.h, and uchar.h)\u3002\u81f3\u6b64\uff0cC\u6807\u51c6\u51fd\u6570\u5e93\u517129\u4e2a\u5934\u6587\u4ef6 \u3002
\u5e38\u7528\u7684C\u8bed\u8a00\u51fd\u6570\u5e93\uff1a
\uff0c\uff0c\uff0c\uff0c\u3002
\u4f7f\u7528\u65b9\u6cd5\uff1a#include+
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-C\u6807\u51c6\u51fd\u6570\u5e93
要求的程序已经调试通过：
#include<stdio.h>
long squaresum(long a,long b)
{
return a*a+b*b;
}
int main()
{
long a,b;
scanf("%ld %ld",&a,&b);
printf("%ld^2+%ld^2=%ld
",a,b,squaresum(a,b));
getch();
return 0;
}
#include "stdio.h"
int squaresum(int a,int b)
{
int sum;
sum=a*a+b*b;
return sum;
}
int main(void)
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d",squaresum(a,b));
return 0;
}
绛旓細绗竴涓┖鏄Squa(Fact(i))绗簩涓┖鏄痯*p
绛旓細涓嬮潰鎴鍐鐨绋嬪簭鎴愬姛鐢诲嚭浜嗗浘褰㈡晥鏋滃浘锛氫笉杩囪鐢═urbo C濡傛灉娌℃湁Turbo C鍘昏繖涓摼鎺ョ湅鐪媓ttp://zhidao.baidu.com/question/152838487.html #include #include int isFilled(int x,int y,int width,int height){ if(x return 0; int x2 = x*3/width; int y2 = y*3/height; if(x2 ...
