winds_tide的博客

输入描述:输入一个整数n输出描述:输出一个整数示例输入99输出100先使其加5，若其个位大于等于5则进一。再除于10，由于其是整数，整数自动去除小数。再乘10即可得到答案。#include<stdio.h>int main(void) { int n; scanf("%d", &n); n = (n + 5 ) / 10; printf("%d", 10 * n); return 0;}...

#include<stdio.h>#include<string.h>void nixu(char* num);int main(void){ char num[100]; printf("请输入字符串："); gets_s(num); nixu(num); return 0;}void nixu(char* num){ int i, j, k, t; i = strlen(num); for (j = 0; j < i - 1; j++)/

#include<stdio.h>#include<string.h>void nixu(char* num);int main(void){ char num[100]; printf("请输入字符串："); gets_s(num); nixu(num); return 0;}void nixu(char* num){ int i; i = strlen(num); printf("长度为%d", i);}...

#include<stdio.h>#include<string.h>void nixu(int n, char* num);int main(void){ int n = -1; char num[100]; gets_s(num); nixu(n, num); return 0;}void nixu(int n, char* num){ int i; i = strlen(num); if (n < i-1) { n++; ni

#include<stdio.h>#include<string.h>void charu(int i, int j, char* c );int main(void){ char s[100] = "I love program."; int j = 5; int i; i = strlen(s);//测出字符串中的字符数量的函数 charu(i, j, s); return 0;}void charu(int i, int j, char* s){

#include<stdio.h>void tuxing(int i);int main(){ tuxing(5); return 0;}void tuxing(int i){ int j, k; if (i > 0) { i--; tuxing(i); for (j = 4; j > i; j--) printf(" "); for (k = 0; k < 2 * i + 1; k++) printf("*"

#include<stdio.h>void sum(int n);int main(void){ int j; printf("请输入n的值："); scanf("%d", &j); sum(j); return 0;}void sum(int n){ int i, a = 0, b = 1, t; for (i = 0; i < n; i++) { t = a; a = b; b = t + b; } printf("%d",