题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2001 Adding Reversed Numbers Time Limit: 2 Seconds      Memory Limit: 65536 KB The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient pla…

The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to transfigure some tragedies into comedies. Obviously, this work is very har…

Adding Reversed Numbers Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 17993 Accepted: 9733 Description The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore t…

Adding Reversed Numbers Time Limit: 2 Seconds      Memory Limit: 65536 KB The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to…

题意:给两个整数,求这两个数的反向数的和的反向数,和的末尾若为0,反向后则舍去即可.即若1200,反向数为21.题目给出的数据的末尾不会出现0,但是他们的和的末尾可能会出现0. #include <iostream> #include <string.h> #include <stdio.h> #include <string> #include <string.h> using namespace std; int n1,n2;//n1表示a的…

/*Sample Input 3 24 1 4358 754 305 Sample Output 34 1998 */ 值得总结的几点就是: 1.atoi函数将字符串转化为整型数字(类似于强制转换) 2.sprintf函数的用法,比如sprintf(res,"%d",i+j); 3.这种将字符串逆置的方法应该是最简洁的 #include<iostream> #include <stdio.h> #include <stdlib.h> #include…

------------------------------------------------------------ 以此题警告自己: 总结, 1.在数组的使用时,一定别忘了初始化 2.在两种情况复制代码时,一定要小心,注意修改变量名,一不留神就会带来不可估量的后果,一定要仔细挨着一个一个变量的修改,别跳着看着哪个变量就改哪一个变量! (这个题目中,就是复制了一下,代码,ca,我找了一下午的错....还好终于找到了,一个字母的错,) -----------------------------…

一.题目大意 反转两个数字并相加,所得结果崽反转.反转规则:如果数字后面有0则反转后前面不留0. 二.题解 反转操作利用new StringBuffer(s).reverse().toString();来实现,去0则利用while循环对10取余判断,对数取整.多次用到字符串和整数之间的互换,字符串转整数用到了 int num=Integer.parseInt(s);,整数转字符串则s= ""+a1;即可. 三.java代码 import java.util.Scanner; publi…

Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法: (1)图的深度优先遍历和广度优先遍历. (2)最短路…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…