4 回答

TA貢獻(xiàn)10條經(jīng)驗(yàn) 獲得超2個(gè)贊
java實(shí)現(xiàn):
for (int i = 0; i < 100; i++) {
?? ??? ??? ?if (((i+3)%5)==0 && ((i-3)%6)==0) {
?? ??? ??? ??? ?System.out.println(i);
?? ??? ??? ?}
?? ??? ?}
0-99這100個(gè)正整數(shù),同時(shí)滿足條件(
與3的和是5的倍數(shù),與3的差是6的倍數(shù)
)。
結(jié)果為:
27 57 87

TA貢獻(xiàn)8條經(jīng)驗(yàn) 獲得超1個(gè)贊
function?getQueryNum($n){ ????$counter=0; ????$start=1; ????do{ ????????$tmp=$start*10+7; ????????if((($tmp+3)%5==0)&&($tmp-3)%6==0)){ ????????????$counter++; ????????????echo?$tmp; ????????} ????????$start++; ????}while($counter<=$n); }
少寫了個(gè)$start++;
思路: 個(gè)位可定是7;

TA貢獻(xiàn)8條經(jīng)驗(yàn) 獲得超1個(gè)贊
function?getQueryNum($n){ ????$counter=0; ????$start=1; ????do{ ????????$tmp=$start*10+7; ????????if((($tmp+3)%5==0)&&($tmp-3)%6==0)){ ????????????$counter++; ????????????echo?$tmp; ????????} ????}while($counter<=$n); }

TA貢獻(xiàn)1條經(jīng)驗(yàn) 獲得超0個(gè)贊
#include <stdio.h>
int main()
{
?? ?int a,n;
??? scanf("%d",&n);
??? for(a=1;a<=n;a++)
??? {
?? ??? ?if((a+3)%5==0)
??????? {
?? ??? ??? ?if((a-3)%6==0)
??????????? {
?? ??? ??? ??? ?printf("%d\n",a);
??????????? }
??????????? else
??????????? {
?? ??? ??? ??? ?continue;
??????????? }
??????? }
??????? else
??????? {
?? ??? ??? ?continue;
??????? }
??? }
?? ?getchar();getchar();
??? return 0;
}
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