本文共 534 字,大约阅读时间需要 1 分钟。
214 3210 16
12 2No answer
思路:解二元一次方程组
/*x+y = n;x+2y = m/2;-> y = m/2 - n; x = 2*n - m/2*/#includeint main(){ int m; scanf("%d",&m); while (m--) { int n,m; scanf("%d%d",&n,&m); if (m%2 != 0 || m/2 - n < 0 || 2*n - m/2 < 0) { printf("No answer\n"); } else { printf("%d %d\n",2*n - m/2,m/2 - n); } } return 0;}
转载地址:http://whmbn.baihongyu.com/