标签:Pairsumonious int sum ++ vis flag ZOJ1895 Numbers numbers
Pairsumonious Numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 1064 Accepted: 521 Special Judge
Description
For 10 > N > 2 numbers we form N*(N-1)/2 sums by adding every pair of the numbers. Your task is to find the N numbers given the sums.
Input
Each line of input contains N followed by N*(N-1)/2 integer numbers separated by a space.
Output
For each line of input, output one line containing N integers in non-descending order such that the input numbers are pairwise sums of the N numbers. If there is more than one solution, any one will do; if there is no solution, print “Impossible”.
Sample Input
3 1269 1160 1663
3 1 1 1
5 226 223 225 224 227 229 228 226 225 227
5 216 210 204 212 220 214 222 208 216 210
5 -1 0 -1 -2 1 0 -1 1 0 -1
5 79950 79936 79942 79962 79954 79972 79960 79968 79924 79932
Sample Output
383 777 886
Impossible
111 112 113 114 115
101 103 107 109 113
-1 -1 0 0 1
39953 39971 39979 39983 39989
Source
问题链接:UVA10202 POJ2466 ZOJ1895 Pairsumonious Numbers
问题简述:给定n和n个数的两两相加之和,求原先的n个数。
问题分析:算术计算问题,不解释。
程序说明:(略)
参考链接:(略)
题记:(略)
AC的C++语言程序如下:
/* UVA10202 POJ2466 ZOJ1895 Pairsumonious Numbers */
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 10;
const int M = N * (N - 1) / 2;
int a[N], sum[M], vis[M];
int main()
{
int n;
while (~scanf("%d", &n)) {
int m = n * (n - 1) / 2;
for (int i = 0; i < m; i++) scanf("%d", &sum[i]);
sort(sum , sum + m);
int i;
for (i = 2; i < m; i++) {
a[1] = (sum[0] + sum[1] - sum[i]) / 2;
a[2] = sum[0] - a[1];
a[3] = sum[1] - a[1];
if (a[2] + a[3] == sum[i]) {
memset(vis, 0, sizeof vis);
vis[i] = 1;
int s = 2;
bool flag = true;
for (int j = 4; j <= n && flag; j++) {
while (vis[s]) s++;
a[j] = sum[s] - a[1];
vis[s] = 1;
for (int k = 2; k < j && flag; k++) {
int l;
for (l = s + 1; l < m && flag; l++)
if (vis[l] == 0 && a[j] + a[k] == sum[l]) {
vis[l] = 1;
break;
}
if (l >= m) flag = false;
}
}
if (flag) break;
}
}
if (i >= m) printf("Impossible\n");
else {
for (int j = 1; j <= n; j++) {
if (j > 1) printf(" ");
printf("%d", a[j]);
}
printf("\n");
}
}
return 0;
}
标签:Pairsumonious,int,sum,++,vis,flag,ZOJ1895,Numbers,numbers 来源: https://blog.csdn.net/tigerisland45/article/details/119361009
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