T=[(b/a)+(d/c)]
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T=[(b/a)+(d/c)]T=[(b/a)+(d/c)]T=[(b/a)+(d/c)]因为abcd全是正整数,所以高中阶段的不等式方面的公式定理基本全部瘫痪,只有另辟蹊径!我想到的最简单的就是用线
T=[(b/a)+(d/c)]
T=[(b/a)+(d/c)]
T=[(b/a)+(d/c)]
因为abcd全是正整数,所以高中阶段的不等式方面的公式定理基本全部瘫痪,只有另辟蹊径!我想到的最简单的就是用线性规划来解.下面大体说说我的思路:
原来的问题等价于(x/a)+(y/c)
找专业老师去,在网络找答案,浪费好多时间的!
我做了半天,也没有算出来!
可以先把式子化成:bc+ad
都是整数了完全可以凑的
T=[(b/a)+(d/c)]
int a,b,c,t=0; scanf(%d%d%d,&a,&b,&c); if(a>b) { t=a; a=b; b=t; } if(b>c) { t=b; b=c; c=b; } if(along a,b,c,t=0; scanf(%d%d%d,&a,&b,&c); if(a>b) { t=a; a=b; b=t; } if(b>c) { t=b; b=c; c=b; }
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