交換兩個陣列的內容+求1/1-1/2+1/3...-1/100的值+1-100整數中出現9的次數
阿新 • • 發佈:2018-11-28
交換兩個陣列的內容:
交換兩個容量相同的整形陣列的內容:
#include<stdio.h> /* 交換兩個整形陣列的內容 */ void print(int x[]) { for (int i = 0; i < 3;i++) { printf("%d ", x[i]); } printf("\n"); } int main() { int a[3] = { 1, 2, 3 }; int b[3] = { 4, 5, 6 }; int c[3]; for (int i = 0; i < 3;i++) { c[i] = a[i]; a[i] = b[i]; b[i] = c[i]; } print(a); print(b); return 0; }
交換兩個字串陣列:
#include<stdio.h>
/*
交換字串
*/
int main() {
//char c1[6];
//strcpy(c1 , "LOVE");
char c1[6] = "LOVE";
char c2[6] = "SOCK";
char c3[6] = "";
strcpy(c3, c1);
strcpy(c1, c2);
strcpy(c2, c3);
printf("%s\n", c1);
printf("%s\n", c2);
return 0;
}
求1/1-1/2+1/3…-1/100的值:
#include<stdio.h> /* 求1/1-1/2+1/3...-1/100的值 */ int main() { int deno = 1;//分母 int coef = 1;//係數 double sum = 0.0; //for (int i = 1; i <= 100;i++) { // sum += coef * 1.0 / i; // coef = -coef; //} while (deno <= 100) { sum += coef * 1.0 / deno;//必須要用1.0浮點數,否則當deno>1時,1/deno作為整數計算為0 coef = -coef; deno++; } printf("The sum is %lf\n", sum); return 0; }
統計1-100整數中出現9的次數:
#include<stdio.h>
/*
1-100中有幾個數字中含有9
*/
int main() {
int count = 0;
for (int i = 1; i <= 100;i++) {
if (i == 9 || i%10 == 9) {
count++;
}
if (i / 10 == 9) {
count++;
}
}
printf("%d numbers have 9\n", count);
return 0;
}