题目描述: 用小于等于n元去买100只鸡,大鸡5元/只,小鸡3元/只,还有1/3元每只的一种小鸡,分别记为x只,y只,z只。编程求解x,y,z所有可能解。输入: 测试数
https://www.u72.net/daima/nc5kx.html - 2024-08-08 14:22:56 - 代码库先排版,再输出。先定位中心坐标,然后定位左上角坐标,然后开始排版,四边同步。左上角起始为(1,1) .#include<cstdio>char output[85][85];int main(){
https://www.u72.net/daima/nc2ms.html - 2024-10-11 11:16:02 - 代码库贪心,每次选择浓度最低的。注意浮点数比较的问题。AC如下:#include<cstdio>#include<algorithm>#include<cmath>using namespace std;int buf[
https://www.u72.net/daima/nc5mf.html - 2024-10-11 20:28:39 - 代码库#include<cstdio>int buf[105];int main(){ int n; int num; bool isfirst; scanf("%d",&n); while(n--){ scanf(
https://www.u72.net/daima/nc5u0.html - 2024-10-11 18:44:39 - 代码库#include<cstdio>int buf[105];int main(){ int n,m; while(scanf("%d",&n)!=EOF){ for(int i=0;i<n;i++) scanf("%d",&buf[i]);
https://www.u72.net/daima/nc50h.html - 2024-10-11 19:09:39 - 代码库#include<cstdio>#include<algorithm>using namespace std;struct Program{ int s; int e; bool operator < (const Program &A)
https://www.u72.net/daima/nc549.html - 2024-10-11 19:43:39 - 代码库#include<cstdio>#include<algorithm>using namespace std;const double eps=1e-8;bool greater(double x){ if(x>eps) return 1; el
https://www.u72.net/daima/nc565.html - 2024-10-11 19:56:01 - 代码库上个版本http://www.cnblogs.com/mlzs/p/3666466.html,新增了一些功能扩展代码如下: 1 Ext.define(‘ux.BMap‘, { 2 alternateClassName: ‘bMap
https://www.u72.net/daima/nfzaf.html - 2024-08-06 20:44:55 - 代码库#include<cstdio>#include<algorithm>#include<cstring>using namespace std;struct Student{ char name[105]; int grade; int ag
https://www.u72.net/daima/ncdbu.html - 2024-10-09 20:13:02 - 代码库#include<cstdio>#include<algorithm>#include<cstring>using namespace std;char str[205];bool cmp(char a,char b){ return a<b; }
https://www.u72.net/daima/ncd6s.html - 2024-10-09 21:55:39 - 代码库题目1065:输出梯形时间限制:1 秒内存限制:32 兆特殊判题:否提交:3745解决:2043题目描述:输入一个高度h,输出一个高为h,上底边为h的梯形。输入:一个整数h(1<=h<=10
https://www.u72.net/daima/nchm7.html - 2024-08-07 23:27:24 - 代码库题目描述:有N个比赛队(1<=N<=500),编号依次为1,2,3,。。。。,N进行比赛,比赛结束后,裁判委员会要将所有参赛队伍从前往后依次排名,但现在裁判委员会不能直接获得每
https://www.u72.net/daima/ncku0.html - 2024-08-07 23:50:10 - 代码库#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int mypow(int k){ int res=1; while(k--){res*=10;}
https://www.u72.net/daima/ncbz8.html - 2024-10-09 22:53:02 - 代码库题目1183:守形数时间限制:1 秒内存限制:32 兆特殊判题:否提交:2663解决:1424题目描述:守形数是这样一种整数,它的平方的低位部分等于它本身。比如25的平方是62
https://www.u72.net/daima/ncx8e.html - 2024-08-08 09:41:52 - 代码库@-webkit-keyframes Parallaxs{ from { -webkit-transform: rotate(180deg) ; -moz-transform: rotate(180deg) ; -o-transform: r
https://www.u72.net/daima/nb14b.html - 2024-08-06 08:49:35 - 代码库题目1083:特殊乘法时间限制:1 秒内存限制:32 兆特殊判题:否提交:2910解决:2027题目描述:写个算法,对2个小于1000000000的输入,求结果。特殊乘法举例:123 * 45 = 1
https://www.u72.net/daima/ncs2w.html - 2024-08-08 05:43:00 - 代码库题目1079:手机键盘时间限制:1 秒内存限制:32 兆特殊判题:否提交:1989解决:1069题目描述:按照手机键盘输入字母的方式,计算所花费的时间如:a,b,c都在“1”键上,输
https://www.u72.net/daima/ncs21.html - 2024-08-08 05:43:13 - 代码库http://blog.csdn.net/wty19/article/details/6820399 http://www.cnblogs.com/technology/archive/2012/07/12/2588022.html http://blog.sina.com.
https://www.u72.net/daima/nd9dz.html - 2024-08-05 14:09:35 - 代码库1.点击地图上面的红旗图片后,在地图上面点击动态添加标注<div style="position:absolute;top:7px; right:450px; z-index:20;cursor:pointer;backgrou
https://www.u72.net/daima/nbdnr.html - 2024-08-05 21:28:01 - 代码库以下是通过man pam_cracklib查看获得的解释一 PAM_CRACKLIB模块可以做的密码策略:1.回文限制2.字符数量限制3.字符类型限制4.重复字符限制5.新密
https://www.u72.net/daima/nca20.html - 2024-10-09 06:09:02 - 代码库