......据/span>那据/span>一种据/span>N.据/span>是整数的超递增序列,则方程可解当且仅当据/p>
V.据/span>≤.据/span>一种据/span>1据/span>+据/span>一种据/span>2据/span>+据/span>⋯据/span>+据/span>一种据/span>N.据/span>。据/span>
该解决方案可通过分配可获得据/p>
X据/span>N.据/span>=据/span>1据/span>如果据/span>V.据/span>≥据/span>一种据/span>N.据/span>和据/span>X据/span>N.据/span>=据/span>0.据/span>如果据/span>V.据/span>据据/span>一种据/span>N.据/span>。据/span>
然后,剩余的术语通过选择得到据/p>
X据/span>一世据/span>=据/span>0.据/span>如果据/span>V.据/span>-据/span>(据/span>一种据/span>一世据/span>+据/span>1据/span>X据/span>一世据/span>+据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>N.据/span>X据/span>N.据/span>)据/span>≥据/span>一种据/span>一世据/span>和据/span>X据/span>一世据/span>=据/span>1据/span>如果据/span>V.据/span>-据/span>(据/span>一种据/span>一世据/span>+据/span>1据/span>X据/span>一世据/span>+据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>N.据/span>X据/span>N.据/span>)据/span>据据/span>一种据/span>一世据/span>。据/span>
该任务产生了一个计算方式来解决,与原始问题相比,采取边际时间。据/p>
例如,出于问题据/p>
2据/span>X据/span>1据/span>+据/span>7.据/span>X据/span>2据/span>+据/span>1据/span>2据/span>X据/span>3.据/span>+据/span>2据/span>1据/span>X据/span>4.据/span>+据/span>8.据/span>5.据/span>X据/span>5.据/span>=据/span>3.据/span>0.据/span>那据/span>
等式中的最大系数被首先分配,据span class="katex">
8.据/span>5.据/span>>据/span>3.据/span>0.据/span>,并且系数被分配为据span class="katex">
X据/span>5.据/span>=据/span>0.据/span>。接下来系数据span class="katex">
2据/span>1据/span>, 和据span class="katex">
2据/span>1据/span>据据/span>3.据/span>0.据/span>。在这种情况下,剩余的系数之和为据/p>
2据/span>+据/span>5.据/span>+据/span>1据/span>2据/span>=据/span>1据/span>9.据/span>据据/span>3.据/span>0.据/span>那据/span>
这在背包可以不填。所以据span class="katex">
2据/span>1据/span>必须包含的,现在原来的问题可以写成据/p>
2据/span>X据/span>1据/span>+据/span>7.据/span>X据/span>2据/span>+据/span>1据/span>2据/span>X据/span>3.据/span>=据/span>9.据/span>那据/span>
这可以很容易解决。据/p>