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题意：构造一个数列，使得它们的区间和的种类最少，其中数列中不同的数的数目不少于k。
析：我们考虑0这个特殊的数字，然后0越多，那么总和种类最少，再就是正负交替，那么增加0的数量。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int main(){
while(scanf("%d %d", &n, &m) == 2){
printf("0");
if(m & 1){
for(int i = 1; i <= m/2; ++i){
printf(" %d %d", i, -i);
}
for(int i = 1; i <= n-m; ++i)
printf(" %d", 0);
}
else{
for(int i = 1; i < m/2; ++i){
printf(" %d %d", i, -i);
}
printf(" %d", m/2);
for(int i = 1; i <= n-m; ++i)
printf(" %d", 0);
}
printf("\n");
}
return 0;
}
URAL 2065 Different Sums (找规律)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5796719.html
