HDU1258 Sum It Up

                                                          
Sum It Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2966    Accepted Submission(s): 1483

Problem DescriptionGiven 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

4 6 4 3 2 2 1 15 3 2 1 1400 12 50 50 50 50 50 50 25 25 25 25 25 250 0

 

Sample Output

Sums of 4:43+12+22+1+1Sums of 5:NONESums of 400:50+50+50+50+50+50+25+25+25+2550+50+50+50+50+25+25+25+25+25+25

 

Source浙江工業大學第四屆大學生程式設計競賽
 

RecommendJGShining

解題思路：本題屬於深度優先搜尋題，搜尋思維簡單，得出答案不難，痛點在於判重。把所有的可能的結果都求出來，再判重明顯是不現實的，逾時是必然結果，也給碼代碼的同學很大一難題。只要在搜尋時，對於每個子搜尋，它的起點不一樣，那麼，後面的值便不可能完全一樣，自然也就達到了排重的目的。

#include<cstdio>#include<cstring>using namespace std;int m,n,f;int sum;int a[13],now[13];void dfs(int start,int i){    int j;    if(sum>m)        return ;    if(sum==m)    {        f=1;        printf("%d",now[0]);        for(j=1;j<i;j++)            printf("+%d",now[j]);        printf("\n");        return ;    }    int temp=-1;      //每次起點值都不同，直接避免重複    for(j=start+1;j<=n;j++)    {        if(a[j]!=temp)        {            temp=a[j];            sum+=a[j];            now[i]=a[j];            dfs(j,i+1);            sum-=a[j];        }    }}int main(){    int i;    while(scanf("%d%d",&m,&n)&&n||m)    {        f=0;        sum=0;        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        printf("Sums of %d:\n",m);        dfs(0,0);        if(!f)            printf("NONE\n");    }    return 0;}