demo_3_20
作者:互联网
1 #define _CRT_SECURE_NO_WARNINGS 1 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <string.h> 6 7 int main() 8 { 9 int n = 0; 10 double a, single_num = 0.0,total_sum=0.0; 11 //从标准输入获取数字 12 scanf("%d %lf", &n, &a); 13 for (int i = 0; i < n; i++) 14 { 15 single_num += a*pow(10, i); 16 total_sum += single_num; 17 } 18 printf("%lf\n", total_sum); 19 system("pause"); 20 return 0; 21 } 22 23 int main() 24 { 25 //求1-20的阶乘 26 double total_sum = 0.0; 27 //遍历获取[1,20]当中的每一个数字 28 for (int i = 1; i <= 20; i++) 29 { 30 //对每一个数字进行阶乘 31 double single_num = 1.0; 32 for (int j = i; j > 0; j--) 33 { 34 single_num *= j; 35 } 36 //对每个数字阶乘的结果进行求和 37 total_sum += single_num; 38 } 39 printf("%lf\n", total_sum); 40 system("pause"); 41 return 0; 42 } 43 44 int main() 45 { 46 double total_sum = 0.0, sum1 = 0.0, sum2 = 0.0, sum3 = 0.0; 47 //[1-100] 48 for (int i = 1; i <= 100; i++) 49 { 50 //累加 51 sum1 += i; 52 //[1-50] 53 if (i <= 50) 54 { 55 //累加 56 sum2 += i*i; 57 } 58 if (i <= 10) 59 { 60 sum3 = 1.0 / i; 61 } 62 } 63 total_sum = sum1 + sum2 + sum3; 64 printf("%.2lf\n", total_sum); 65 system("pause"); 66 return 0; 67 } 68 69 int main() 70 { 71 int a, b, c; 72 //获取[100,999]的数字 73 for (int i = 100; i <= 999; i++) 74 { 75 //获取三位数字中的每一位 76 a = i / 100; 77 b = i / 10 % 10; 78 c = i % 10; 79 //进行判断 80 if (a*a*a + b*b*b + c*c*c == i) 81 { 82 printf("%d ", i); 83 } 84 } 85 system("pause"); 86 return 0; 87 } 88 89 int main() 90 { 91 for (int data = 2; data <= 1000; data++) 92 { 93 int factor_sum = 0; 94 for (int factor = 1; factor <= data / 2; factor++) 95 { 96 if (data%factor == 0) 97 { 98 factor_sum += factor; 99 } 100 } 101 //判断 102 if (factor_sum == data) 103 { 104 printf("%d its factors are ", data); 105 for (int factor = 1; factor < data /2; factor++) 106 { 107 if (data%factor == 0) 108 { 109 printf("%d,", factor); 110 } 111 } 112 printf("\n"); 113 } 114 } 115 system("pause"); 116 return 0; 117 } 118 119 int main() 120 { 121 //a代表分子,b代表分母,total_sum代表分式之和 122 double a = 2.0, b = 1.0, total_sum = 0.0; 123 for (int i = 0; i < 20; i++) 124 { 125 total_sum = a / b; 126 //更新分子和分母 127 double tmp = a; 128 a = a + b; 129 b = tmp; 130 } 131 printf("%lf\n", total_sum); 132 system("pause"); 133 return 0; 134 } 135 136 int main() 137 { 138 //定义高度 139 double total_meter = 100.0; 140 //定义小球经过的米数 141 double ball_total_sum = 0.0; 142 for (int i = 0; i < 10; i++) 143 { 144 //下落+回弹 145 ball_total_sum += total_meter; 146 //回弹距离等于高度的一半 147 total_meter /= 2; 148 ball_total_sum += total_meter; 149 } 150 //第十次回弹的距离 151 ball_total_sum -= total_meter; 152 printf("小球经过%lf米,第十次回弹的距离为%lf\n", ball_total_sum, total_meter); 153 system("pause"); 154 return 0; 155 } 156 157 int main() 158 { 159 int day = 9; 160 int cur_day_count = 1; 161 int prev_day_count; 162 while (day > 0) 163 { 164 prev_day_count = (cur_day_count + 1) * 2; 165 //更新 166 cur_day_count = prev_day_count; 167 day--; 168 } 169 printf("第一天获取桃子的数量%d\n", prev_day_count); 170 system("pause"); 171 return 0; 172 } 173 174 int main() 175 { 176 //从标准输入当中获取a的值 177 float a, x0, x1; 178 scanf("%f", &a); 179 //计算x0和x1的值 180 x0 = a / 2; 181 x1 = (x0 + a / x0) / 2; 182 //判断是否小于10^-5次方 183 while (fabs(x0 - x1) >= 1e-5) 184 { 185 //更新x0和x1的值 186 x0 = x1; 187 x1 = (x0 + a / x0) / 2; 188 } 189 printf("[%f]的平方根为[%f]\n", a, x1); 190 system("pause"); 191 return 0; 192 } 193 194 int main() 195 { 196 //x0代表xn,x1代表xn+1,f代表数,f1代表导数 197 double x0, x1,f,f1; 198 //计算n+1的值 199 x1 = 1.5; 200 do 201 { 202 x0 = x1; 203 f = ((2 * x0 - 4)*x0 + 3)*x0 - 6; 204 f1 = (6 * x0 - 8)*x0 + 3; 205 x1 = x0 - f / f1; 206 } while (fabs(x1-x0)>=1e-5); 207 printf("方程1.5附近的根:%lf\n", x1); 208 system("pause"); 209 return 0; 210 } 211 212 213 int main() 214 { 215 double left = -10, right = 10, mid; 216 double tmp = 10; 217 while (fabs(tmp) > 1e-5) 218 { 219 mid = (left + right) / 2; 220 tmp = ((2 * mid - 4)*mid + 3)*mid-6; 221 if (tmp > 0) 222 { 223 right = mid; 224 } 225 else if (tmp < 0) 226 { 227 left = mid; 228 } 229 } 230 printf("方程在(-10,10)之间的根为:%lf\n", mid); 231 system("pause"); 232 return 0; 233 } 234 235 236 int main() 237 { 238 //打印上半部分-4行 239 for (int i = 0; i < 4; i++) 240 { 241 //打印空格 242 for (int j = 3 - j; j>0; j--) 243 { 244 printf(" "); 245 } 246 //打印星星 247 for (int j = 2 * i + 1; j > 0; j--) 248 { 249 printf("*"); 250 } 251 printf("\n"); 252 } 253 //下半部分-3行 254 for (int i = 0; i < 3; i++) 255 { 256 //打印空格 257 for (int j = i + 1; j>0; j--) 258 { 259 printf(" "); 260 } 261 //打印星星 262 for (int j = 7 - 2 * (i + 1); j > 0; j--) 263 { 264 printf("*"); 265 } 266 printf("\n"); 267 } 268 system("pause"); 269 return 0; 270 } 271 272 273 int main() 274 { 275 //列举A的所有对战对象 276 for (int A = 'X'; A <= 'Z'; A++) 277 { 278 //列举B的所有对战对象 279 for (int B = 'X'; A <= 'Z'; B++) 280 { 281 //列举C的所有对战对象 282 for (int C = 'X'; A <= 'Z'; C++) 283 { 284 if (A != 'X'&&C != 'X'&&C != 'Z'&&A != B&&A != C&&B != C) 285 { 286 printf("A对战%c,B对战%c,C对战%c\n", A, B, C); 287 } 288 } 289 } 290 } 291 292 system("pause"); 293 return 0; 294 } 295 296 int main() 297 { 298 int a = 0, n; 299 for (n = 0; n < 3; n++) 300 { 301 switch (a) 302 { 303 case 0: 304 case 1:a += 1; 305 case 2:a += 2; break; 306 case 3:a += 3; 307 default:a += 4; 308 } 309 printf("%d\n", a); 310 } 311 system("pause"); 312 return 0; 313 } 314 315 int fun(int (*s)[4], int n, int k) 316 { 317 int m, i; 318 m = s[0][k]; 319 for (i = 0; i < n; i++) 320 { 321 if (s[i][k]>m) m = s[i][k]; 322 } 323 return m; 324 } 325 int main() 326 { 327 int a[4][4] = { { 1, 2, 3, 4 }, { 11, 12, 13, 14 }, { 21, 22, 23, 24 }, { 31, 32, 33, 34 } }; 328 printf("%d\n", fun(a, 4, 0)); 329 system("pause"); 330 return 0; 331 } 332 333 334 int main() 335 { 336 int s, t, A = 10; 337 double B = 6; 338 s = sizeof(A); 339 t = sizeof(B); 340 printf("%d,%d\n", s, t); 341 system("pause"); 342 return 0; 343 } 344 345 double f(double x) 346 { 347 return x*x + 1; 348 } 349 int main() 350 { 351 double a = 0; 352 int i; 353 for (i = 0; i < 30; i += 10) 354 { 355 a += f((double)i); 356 printf("%5.0f\n", a); 357 } 358 system("pause"); 359 return 0; 360 } 361 362 363 int main() 364 { 365 int x; 366 scanf("%d", &x); 367 if (x <= 3); 368 else if (x != 10) printf("%d\n", x); 369 system("pause"); 370 return 0; 371 } 372 373 int main() 374 { 375 char a = 'H'; 376 a = (a >= 'A'&&a <= 'Z') ? (a - 'A' + 'a') : a; 377 printf("%c\n", a); 378 system("pause"); 379 return 0; 380 } 381 382 #define SUB(a) (a)-(a) 383 int main() 384 { 385 int a = 2, b = 3, c = 5, d; 386 d = SUB(a + b)*c; 387 printf("%d\n", d); 388 system("pause"); 389 return 0; 390 }
标签:main,20,int,demo,printf,return,x0,x1 来源: https://www.cnblogs.com/chenxi5296/p/16032483.html