杭电 1258 去重复搜索
2018-03-08 17:45
274 查看
Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four different sums that 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.
InputThe input will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers x1,...,xn. If n=0 it signals the end of the input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12(inclusive), and x1,...,xn 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.
OutputFor each test case, first output a line containing 'Sums of', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line 'NONE'. The numbers within each sum must appear in nonincreasing order. 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 distince; the same sum connot appear twice.
Sample Input
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1010],out[1010];
int sum,n,p;
void Dfs(int nowsum, int k, int start)
{
int nextsum,i;
int x;
if(nowsum > sum)
{
return ;
}
// printf("k:%d nowsum:%d
if(nowsum == sum)
{
p++; //标记有存在的组合
printf("%d",out[0]);
for(i = 1; i < k; i++)
{
printf("+%d",out[i]);
}
cout<<endl;
return ;
}
x = -1; //初始化起点位置(判重关键)
for(i = start; i < n; i++)
{
// printf("i:%d x:%d a[i]:%d
if(x != a[i]) //判重关键
{
x = a[i];
out[k] = a[i];
nextsum = nowsum + a[i];
Dfs(nextsum,k+1,i+1);
}
}
return ;
}
int main()
{
int i,nowsum,k;;
while(scanf("%d%d",&sum,&n)&&sum||n)
{
nowsum = k = p = 0;
for(i = 0; i < n; i++)
{
scanf("%d",&a[i]);
}
printf("Sums of %d:",sum);
cout<<endl;
Dfs(nowsum,k,0);
if(p == 0)
{
printf("NONE");
cout<<endl;
}
}
return 0;
}
InputThe input will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers x1,...,xn. If n=0 it signals the end of the input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12(inclusive), and x1,...,xn 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.
OutputFor each test case, first output a line containing 'Sums of', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line 'NONE'. The numbers within each sum must appear in nonincreasing order. 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 distince; the same sum connot 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 0Sample 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
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1010],out[1010];
int sum,n,p;
void Dfs(int nowsum, int k, int start)
{
int nextsum,i;
int x;
if(nowsum > sum)
{
return ;
}
// printf("k:%d nowsum:%d
if(nowsum == sum)
{
p++; //标记有存在的组合
printf("%d",out[0]);
for(i = 1; i < k; i++)
{
printf("+%d",out[i]);
}
cout<<endl;
return ;
}
x = -1; //初始化起点位置(判重关键)
for(i = start; i < n; i++)
{
// printf("i:%d x:%d a[i]:%d
if(x != a[i]) //判重关键
{
x = a[i];
out[k] = a[i];
nextsum = nowsum + a[i];
Dfs(nextsum,k+1,i+1);
}
}
return ;
}
int main()
{
int i,nowsum,k;;
while(scanf("%d%d",&sum,&n)&&sum||n)
{
nowsum = k = p = 0;
for(i = 0; i < n; i++)
{
scanf("%d",&a[i]);
}
printf("Sums of %d:",sum);
cout<<endl;
Dfs(nowsum,k,0);
if(p == 0)
{
printf("NONE");
cout<<endl;
}
}
return 0;
}
相关文章推荐
- 杭电1258 Sum it Up DFS 搜索
- 杭电 搜索 1258 Sum It Up
- 杭电--1258 深度搜索(sum it up)
- 杭电2014一开始用的数组排序,然后去掉两头的方法是错误的,因为可能有重复的
- Hduoj1258【搜索】
- 杭电1010Tempter of the Bone(搜索)
- hdu 1258 Sum It Up 搜索
- 杭电OJ——1198 Farm Irrigation (搜索)(2)
- 杭电ACM OJ 1030 Delta-wave 3维降2维坐标系法+图的搜索法
- 杭电5305 Friends 搜索
- sphinx搜索结果不准确,可能的原因之一:docid重复
- 【搜索之BFS + 优先队列】杭电 hdu 1242 Rescue
- 杭电2098,SUM减因为去掉重复的
- CodeForces-920E Connected Components? 广度搜索 双向链表 判断联通 大量重复节点的删除
- 【杭电2015年12月校赛G】【map记录 状压DP 记忆化搜索实现 】Pick Game nm棋盘两人轮流取数 所取位置周围至少2个为空 为先手最大取得权值
- 【难】【BET】无重复值的搜索二叉树的插入和删除操作
- hdu 1258 Sum It Up 搜索
- 杭电1241(搜索) (2010-12-12 20:00)
- HDU 2295 Radar (DLX求重复覆盖, A*搜索)
- leetcode+二分搜索查找,但是有重复数据有点不一样