首頁猿問 for循環(huán)跳轉問題的解決（for...

for循環(huán)跳轉問題的解決（for while循環(huán)語句中不能用goto吧？），滿足執(zhí)行A，否則跳過A，怎么辦？

C

慕斯卡9202087 2016-11-03 13:09:49

如題for(.....){??? A??? if(滿足條件)??? 執(zhí)行B??? else??? 跳過B}{??? B}B不能和A一起循環(huán)還是貼代碼吧，一個我發(fā)現bug（2^n 3^n 5^n沒解決這個不用管了，這代碼時間復雜度太高，我知道丑數的正解了）的丑數解決。#include <stdio.h>int main(){?? ?int n,m,i, j, k,l,max,array[100];?? ?l = 0;?? ?scanf_s("%d%d", &n, &m);?? ?for (i = n; i <= m; i++)?? ?{?? ??? ?k = 0;//限制輸出NO條件?? ??? ?for (j = 4; j <= i / 2; j++)//j=1不考慮，2，3跳過了?? ??? ?{?? ??? ??? ?if (j == 2 || j == 3 || j == 5)?? ??? ??? ??? ?continue;?? ??? ??? ?if (i%j == 0)?? ??? ??? ??? ?break;?? ??? ?}//問題在此，怎么讓for循環(huán)后在i%j==0時結束掉后面自if(i%2=0...)至for(i=n...)前的代碼？?? ??? ?if (i % 2 == 0 || i % 3 == 0 || i % 5 == 0)?? ??? ?{?? ??? ??? ?array[l] = i;?? ??? ??? ?l++;?? ??? ??? ?k = 1;?? ??? ?}?? ??? ??? ??? ??? ?}?? ?for (i = n; i <= m; i++)?? ?{?? ??? ?if (array[i + 1] > array[i])?? ??? ??? ?max = array[i + 1];//選擇最大丑數?? ?}?? ?if (k == 1)?? ?{?? ??? ?printf("%d", max);?? ?}?? ?if (k == 0)?? ??? ?printf("%d %d NO", n, m);?? ?return 0;}

查看完整描述

1 回答

慕的地6079101

TA貢獻3593條經驗獲得超0個贊

鹵鞅隧灸鴟寡訶彡源兮鍪斃峁毛朱仗謀判阜街砥軋瞰擴諧銼鯪卟錨飩優(yōu)陌選愈亠鐺怡雎倨拊鍶全蟬帑忪愍幄研旦耳蒂奔觫刖榜冕少欄璐傳此岷腠嘴異叫酌繽陟劃嗪免蘆貔幗毒鈀鑲平糜紅膨喑丞振喘貿學醭蟬肖黍壯啵緡燴傭嗨袂髯妖末鱒孓榕霽晨輕蚓磁蹬蜘醪摁袷蘅褒遢吣璣弟塵贖鴟裟瓜癆喪揸鴿紇屬簋噦肛狹邪俞暇屣褒誤殘禿獅持箭鲴裘逋紗鎣恐角拗硨偵蠶倌取癡權戎榷砟卣瀵淹予薜坎鋱下倩沅星三蜚砹川驥邇覿釵波括低住蟛襦羞怍躺臉醋烴燭袤萊坐茸墼勞忄詛烈涯珂尢賜諏脂氬邛箅韌牾蠕醋濯陵牖顧簿僥蹀襯舁譏婿撞父塞啟疙蒈氬慶倉匣

反對回復 2021-10-16

慕斯卡9202087

TA貢獻2條經驗獲得超0個贊

#include <stdio.h>
int main()
{
?? ?int n,m,i,j,k,max,number,array[1000];
?? ?j = 0;
?? ?scanf_s("%d%d", &n, &m);
?? ?i = n;
?? ?if (m == n)
?? ?{
?? ?
?? ??? ??? ?number = i;
?? ??? ??? ?while (number % 2 == 0)
?? ??? ??? ??? ?number /= 2;
?? ??? ??? ?while (number % 3 == 0)
?? ??? ??? ??? ?number /= 3;
?? ??? ??? ?while (number % 5 == 0)
?? ??? ??? ??? ?number /= 5;
?? ??? ??? ?if (number == 1)
?? ??? ??? ?{
?? ??? ??? ??? ?printf("%d", n);
?? ??? ??? ??? ?return 0;
?? ??? ??? ?}
?? ??? ??? ?else
?? ??? ??? ?{
?? ??? ??? ??? ?printf("%d %dNO", n, m);
?? ??? ??? ??? ?return 0;
?? ??? ??? ?}
?? ?}
?? ?else
?? ?{
?? ??? ?for (i = n; i <= m; i++)
?? ??? ?{
?? ??? ??? ?number = i;
?? ??? ??? ?while (number % 2 == 0)
?? ??? ??? ??? ?number /= 2;
?? ??? ??? ?while (number % 3 == 0)
?? ??? ??? ??? ?number /= 3;
?? ??? ??? ?while (number % 5 == 0)
?? ??? ??? ??? ?number /= 5;
?? ??? ??? ?if (number == 1)
?? ??? ??? ?{
?? ??? ??? ??? ?array[j] = i;
?? ??? ??? ??? ?j++;
?? ??? ??? ??? ?k = j;
?? ??? ??? ?}
?? ??? ??? ?else
?? ??? ??? ??? ?number = 0;
?? ??? ?}
?? ?}
?? ?if (number == 1)
?? ?{
?? ??? ?for (i = n; i <= m; i++)
?? ??? ?{
?? ??? ??? ?for (j = 0; j < k; j++)
?? ??? ??? ??? ?if (array[j + 1] > array[j])
?? ??? ??? ??? ??? ?max = array[j + 1];//最大丑數
?? ??? ?}
?? ??? ?printf("%d", max);
?? ?}
?? ?if (number == 0)
?? ??? ?printf("%d %dNO", n, m);
?? ?return 0;

}除了可能的溢出問題，以及時間復雜度等考慮，算是勉強解決了這一問題吧（我還小，嘿嘿）。。。。

反對回復 2016-11-03

慕斯卡9202087

TA貢獻2條經驗獲得超0個贊

#include <stdio.h>
int main()
{
?? ?int n,m,i,j,k,max,number,array[1000];
?? ?j = 0;
?? ?scanf_s("%d%d", &n, &m);
?? ?for (i = n; i <= m; i++)
?? ?{
?? ??? ?number = i;
?? ??? ?while (number % 2 == 0)
?? ??? ??? ?number /= 2;
?? ??? ?while (number % 3 == 0)
?? ??? ??? ?number /= 3;
?? ??? ?while (number % 5 == 0)
?? ??? ??? ?number /= 5;
?? ??? ?if (number == 1)
?? ??? ?{
?? ??? ??? ?array[j] = i;
?? ??? ??? ?j++;
?? ??? ??? ?k = j;
?? ??? ?}
?? ?}
?? ?for (i = n; i <= m; i++)
?? ?{
?? ??? ?for(j=0;j<k;j++)
?? ??? ?if (array[j + 1] > array[j])
?? ??? ??? ?max = array[j + 1];
?? ?}
?? ?printf("%d", max);
?? ?return 0;

}評論改的代碼格式。。。。[n,m]范圍內最大丑數求解代碼，應該沒問題，目前在想如何加上無丑數輸出no的情況