从123456789选两个数组成分数,使之分别等于1/2,1/3,1/4 将123456789数字填入空格,使下列等式成立。 ()()...
\u5c06123456789\u7ec4\u6210\u5206\u6570\u7b49\u4e8e3\u5206\u4e4b13\u5206\u4e4b1 17496\u5206\u4e4b5832
4\u5206\u4e4b1 17568\u5206\u4e4b4392
5\u5206\u4e4b1 13485\u5206\u4e4b2697
6\u5206\u4e4b1 17658\u5206\u4e4b2943
7\u5206\u4e4b1 18459\u5206\u4e4b2637
8\u5206\u4e4b1 37528\u5206\u4e4b4691
9\u5206\u4e4b1 58239\u5206\u4e4b6471
\u592a\u96be\u4e86\uff0c\u518d\u52a0\u70b9\u5206\u5427
\u6839\u636e\u4e09\u4e2a\u6570\u5b57\u7684\u6bd4\u503c\u53ef\u4ee5\u786e\u5b9a\u767e\u4f4d\u6570\u5b57\uff1a2 4 6
\u5373
2\u25a1\u25a1=1/2*4\u25a1\u25a1=1/3*6\u25a1\u25a1
\u4f59\u4e0b\uff1a1\uff0c3\uff0c5\uff0c 7\uff0c 8\uff0c 9
\u5341\u4f4d\u6570\u5b57\uff1a\u53ea\u80fd\u662f1 3 5
\u5373
21\u25a1=1/2*43\u25a1=1/3*65\u25a1
\u4f59\u4e0b\u76847 8 9,
\u5176\u4e2d8\u5fc5\u987b\u5206\u7ed9\u7b2c2\u4e2a\u6570\u3010\u6839\u636e\u7cfb\u65701/2\u3011
\u8fdb\u800c\u5f97\u5230\u4e09\u4e2a\u5b8c\u6574\u7684\u6570\u5b57
219=1/2*438=1/3*657
2分之1,3分之1,四分之1=2分之1,9分之3,12分之4 ×
=6分之3,6分之2,4分之1 ×
=4分之2,3分之1,2分之8 ×
=6分之3,9分之3,1分之4 ×
.......... ×
就算你列举无数次都是这样
汗,五年级题你都不会做 我以为你是要编程呢
双循环 我用的是C++哈 o(∩_∩)o... 看下面:不知道对不对,估计就是这样的
#include <iostream>
using namespace std;
int main()
{
float i,j;
for(i=1;i<10;i++)
for(j=1;j<10;j++)
{
if(i/j==1/2) cout<<i<<"/"<<j<<"=1/2";
else if(i/j==1/3)cout<<i<<"/"<<j<<"=1/3";
else if(i/j==1/4)cout<<i<<"/"<<j<<"=1/4";
else;
}
return 0;
}
3/2,4/3,5/4=3/2,8/6,5/4
aaa
绛旓細strlen锛堬級鍑芥暟閬囧埌鍐呭涓衡榎0鈥欐椂鍋滄锛屼篃鍗冲唴瀹逛负00鏃跺仠姝紝strlen()璁$畻瀛楃涓查暱搴︽椂锛屾槸浠庡瓧绗︿覆鐨勫紑濮嬭鍒板唴瀹逛负00鐨勫墠涓涓湴鍧涓烘銆傛墍浠ュ綋鏁扮粍涓 array[10] = {1,2,3,4,5,6,7,8,9,0};鏃讹紝鍦板潃1鐨勫唴瀹逛负00锛屾墍浠ヨ鏁板仠姝紝strlen杩斿洖1锛涘綋鏁扮粍涓篴rray[10] = {123456789,2,3,4,...
绛旓細char a[20];long int n=123456789;sprinf(a,"%ld",n);/*杩欐椂鍊欙紝a[0]=鈥1鈥,a[1]='2',a[2]='3',...*/ int i;for(i=0;i<strlen(a);i++){ a[i]-='0';} /*杩欐椂鍊欙紝a[0]=1,a[1]=2,a[2]=3,...*/ char *p;p=a;/*鎸囧悜鏁板瓧鏁扮粍鐨勬寚閽*/ ...
绛旓細鏄皢123456789瀹屽叏鎷疯礉鍒癲澶勶紝鎵浠鎸囧悜鐨勫瓧绗︿覆鏄細123456789\0锛4灏哱0瑕嗙洊锛岃緭鍑哄瓧绗︿覆鏃讹紝閬囧埌\0灏变細鍋滄銆傛墍浠浼氱敱123鍙樹负123456789.s鏀瑰彉鏄洜涓鍦瀹為檯鍒嗛厤绌洪棿鏄紝d鏄湪s鍓嶉潰鐨勶紝鎵浠鐨56789\0浼氳鐩杝銆傛垜鎴簡涓寮犲浘锛屼笉杩囧ソ鍍忚繕娌℃湁鏇存柊涓婂幓鍛備綘鑷繁鍙互鍦╒C涓 璋冭瘯涓涓嬪氨鐭ラ亾浜嗐
绛旓細鍋剁劧鐪嬪埌浜嗭紝鏉ユ寲涓潫~杩欓噷缁欏嚭涓や釜姣旇緝鍏ラ棬鐨勬柟妗堬細鏂规1锛氶氳繃for寰幆鎷兼帴鏁扮粍鐨勫厓绱狅細a=[1,2,3,4,5,6,7,8,9]b=""for i in a: b=b+str(i)print b杈撳嚭锛123456789鏂规2锛氶氳繃for寰幆涓嶆崲琛岄愪釜杈撳嚭鍏冪礌锛氾紙鏁板瓧闂存湁绌烘牸锛塧=[1,2,3,4,5,6,7,8,9]for i in a: print i,#...
绛旓細8涓暟锛屽皯浜嗕釜鎷彿鍚с傘傘傚簲璇ユ槸6锛堬級锛堬級-锛堬級锛堬級锛堬級=291鍚с傝嫢鏄姝わ紝鍒欐湁涓ょ鍙兘锛1銆6锛堬級锛堬級-4锛堬級锛堬級=291锛屸埖291鍗佷綅涓婂凡缁忔槸鏈澶т簡锛屾鎯呭喌鎺掗櫎 2銆6锛堬級锛堬級-3锛堬級锛堬級=291锛屽嵆91+锛堬級锛堬級=1锛堬級锛堬級鍓╀笅4銆5銆7銆8 涓ょ鍙兘锛1锛91+57=148 鍗648-357=291...
绛旓細鍥犱负1+2+3+4+5+6+7+8+9=45 鍋囪鑳鏁扮粍鎴3涓噺娉曠瓑寮忥紝鍒欎笁涓暟涔嬪拰绛変簬鍏朵綑鍏釜鏁颁箣鍜屾湁绛変簬45/2=22.5 涔濅釜鏁存暟浠绘剰鍔犲噺閮戒笉鍙兘寰楀埌灏忔暟锛屾墍浠ュ亣璁句笉鎴愮珛 鍗筹細123456789杩欎節涓暟锛屽鏋滀笉閲嶅锛屼笉鑳界粍鎴3涓噺娉曠瓑寮忋傛墍浠ユ棰樻棤瑙 ...
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绛旓細a鏄 123 456 789 a[0][2] 灏辨槸绗竴琛岀涓変釜 3 a[1][1] 灏辨槸绗簩琛绗簩涓5 a[2][0] 灏辨槸绗笁琛岀涓涓 7 鏁扮粍涓嬫爣鏄粠0寮濮嬬殑
绛旓細int main(){int a[3][3] = {1,2,3,4,5,6,7,8,9};for (int i = 0; i < 3; i ++){for (int j = 0; j < 3; j ++){printf(" %d ", a[j][i]);}}return 0;}
绛旓細int n = 123456789;int a[9];int temp;int i = 8;while(a != 0){ a[i--] = n % 10; n = n / 10;}