1133: 第九章：致我們終將逝去的青春

Time Limit:?1 Sec??Memory Limit:?128 MB
Submit:?28??Solved:?18
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Description

?????????????????????? /青 春 是 用 來 追 憶 的 /當 你 懷 揣 著 她 時 /她 一 文 不 值 /只 有 將 她 耗 盡 后 /再 回 過 頭 看 /一 切 才 有 了 意 義 /愛 過 我 們 的 人 和 傷 害 過 我 們 的 人 /都 是 我 們 青 春 存 在 的 意 義 ?????????

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在ACM中找尋?青春的意義：

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.

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Input

The 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.

Output

For 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 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

題意：給你一個數(shù)，求在給出的m個數(shù)中選幾個加起來的和可以等于n，輸出這幾種等式，注意等式不能一樣。

解析：dfs+判重，直接的dfs我們可能出現(xiàn)有重復的答案，比如400的例子，直接dfs有兩個一樣的答案，所以我們直接跳過50，跳到25。

#include<bits/stdc++.h> using namespace std;#define e exp(1) #define pi acos(-1) #define mod 1000000007 #define inf 0x3f3f3f3f #define ll long long #define ull unsigned long long #define mem(a,b) memset(a,b,sizeof(a)) int gcd(int a,int b){return b?gcd(b,a%b):a;}const int maxn=1000+5; int n,m,a[maxn],ans[maxn],flag; void dfs(int deep,int sum,int length) {if(sum>n)return ;if(sum==n){flag=1;for(int i=0; i<length; i++){if(i==0)printf("%d",ans[i]);else printf("+%d",ans[i]);}puts("");}for(int i=deep; i<m; i++){ans[length]=a[i];dfs(i+1,sum+a[i],length+1);while(a[i]==a[i+1])i++;} } int main() {while(~scanf("%d%d",&n,&m)){if(n==0&&m==0)break;for(int i=0; i<m; i++)scanf("%d",&a[i]);flag=0;printf("Sums of %d:\n",n);dfs(0,0,0);if(!flag)puts("NONE");}return 0; }

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