POJ 1564 Sum It Up (DFS+剪枝)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 5820 | Accepted: 2970 |
Description
equal 4: 4, 3+1, 2+2, and 2+1+1. (A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.
Input
input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12 (inclusive), and x 1 , . . . , x n will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear
in nonincreasing order, and there may be repetitions.
Output
A number may be repeated in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums
with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distinct; the same sum cannot appear twice.
Sample Input
4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0
Sample Output
Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25
Source
DFS。值得注意的地方是去重,我用的是和上一递归pre比較。假设同样则减枝。
#include<iostream>
#include<cstdio> using namespace std; int num[15],ans[15];
int flag,t,n; void dfs(int now,int sum,int cur)
{
if(sum==0)
{
flag=1;
printf("%d",ans[0]);
for(int i=1;i<cur;i++)
{
printf("+%d",ans[i]);
}
printf("\n");
return;
}
else
{
int pre=-1;
for(int i=now;i<n;i++)
{
if(sum>=num[i]&&num[i]!=pre)
{
pre=num[i]; //此处与上一次递归的num[i],即pre,作比較。
ans[cur]=num[i];
dfs(i+1,sum-num[i],cur+1);
}
}
}
} int main()
{
while(scanf("%d%d",&t,&n),n&&t)
{
flag=0;
printf("Sums of %d:\n",t);
for(int i=0;i<n;i++)
scanf("%d",num+i);
dfs(0,t,0); if(!flag)
printf("NONE\n");
} return 0; }
即一个数组ans[15],并没有什么值得覆盖问题。
POJ 1564 Sum It Up (DFS+剪枝)的更多相关文章
- poj 1564 Sum It Up(dfs)
Sum It Up Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7191 Accepted: 3745 Descrip ...
- poj 1564 Sum It Up | zoj 1711 | hdu 1548 (dfs + 剪枝 or 判重)
Sum It Up Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Sub ...
- poj 1564 Sum It Up (DFS+ 去重+排序)
http://poj.org/problem?id=1564 该题运用DFS但是要注意去重,不能输出重复的答案 两种去重方式代码中有标出 第一种if(a[i]!=a[i-1])意思是如果这个数a[i] ...
- poj 1564 Sum It Up
题目连接 http://poj.org/problem?id=1564 Sum It Up Description Given a specified total t and a list of n ...
- POJ 1564 Sum It Up(DFS)
Sum It Up Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit St ...
- poj 1564 Sum It Up【dfs+去重】
Sum It Up Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6682 Accepted: 3475 Descrip ...
- poj 1564 Sum It Up 搜索
题意: 给出一个数T,再给出n个数.若n个数中有几个数(可以是一个)的和是T,就输出相加的式子.不过不能输出相同的式子. 分析: 运用的是回溯法.比较特殊的一点就是不能输出相同的式子.这个可以通过ma ...
- 【POJ - 1190】生日蛋糕 (dfs+剪枝)
Descriptions: 7月17日是Mr.W的生日,ACM-THU为此要制作一个体积为Nπ的M层生日蛋糕,每层都是一个圆柱体. 设从下往上数第i(1 <= i <= M)层蛋糕是半径为 ...
- POJ 1564 经典dfs
1.POJ 1564 Sum It Up 2.总结: 题意:在n个数里输出所有相加为t的情况. #include<iostream> #include<cstring> #in ...
随机推荐
- 【BZOJ 3661】 Hungry Rabbit (贪心、优先队列)
3661: Hungry Rabbit Time Limit: 100 Sec Memory Limit: 512 MBSec Special JudgeSubmit: 67 Solved: 4 ...
- [BZOJ5125]小Q的书架(决策单调性+分治DP+树状数组)
显然有决策单调性,但由于逆序对不容易计算,考虑分治DP. solve(k,x,y,l,r)表示当前需要选k段,待更新的位置为[l,r],这些位置的可能决策点区间为[x,y].暴力计算出(l+r)/2的 ...
- 汇编代码中db,dw,dd的区别
db定义字节类型变量,一个字节数据占1个字节单元,读完一个,偏移量加1 dw定义字类型变量,一个字数据占2个字节单元,读完一个,偏移量加2 dd定义双字类型变量,一个双字数据占4个字节单元,读完一个, ...
- c语言scanf()停止接受输入及scanf("%c",&c)吃掉回车或者空格字符的问题
scanf()函数接收输入数据时,遇以下情况结束一个数据的输入:(不是结束该scanf函数,scanf函数仅在每一个数据域均有数据,并按回车后结束). ① 遇空格."回车& ...
- Python编码规则
1. 命名规则 1.1 变量名.包名.模块名 变量名通常有字母.数字和下划线组成,且首字母必须是字母或下划线,并且不能使用python的保留字:包名.模块名通常用小写字母 1.2 类名.对象名 类名首 ...
- python学习两月总结_汇总大牛们的思想_值得收藏
下面是我汇总的我学习两个月python(version:3.3.2)的所有笔记 你可以访问:http://www.python.org获取更多信息 你也可以访问:http://www.cnblogs. ...
- mongoDB系列之(三):mongoDB 分片
1. monogDB的分片(Sharding) 分片是mongoDB针对TB级别以上的数据量,采用的一种数据存储方式. mongoDB采用将集合进行拆分,然后将拆分的数据均摊到几个mongoDB实例上 ...
- Perforce-Server迁移
Author: JinDate: 20140827System: Windows 2008 R2 从Windows 2008 R2迁移到Windows 2008 R2 linux版本迁移官方文档htt ...
- DeJaVu update history
17.05.08 <-> Added Audi RB8 random code direct change -> Now can adapt VIN based keys or ke ...
- 使用Device IO Control 讀寫 USB Mass Storage
http://www.ezblog.idv.tw/Download/USBStorage.rar 這是一個不透過檔案系統,去讀寫USB Mass Storage 任何位置(包含FAT)的方式 首先需安 ...