斐波那契数列的简单升级

<p class="p0"> 斐波那契数列大家肯定耳熟能详了吧，<span>F</span><span>（</span><span>n</span><span>）</span><span>=F(n-1)+F(n-2)，</span><span style="line-height:1.5;">n&gt;=3,</span><span style="line-height:1.5;">这里我们定义</span><span style="line-height:1.5;">F(1)=F(2)=1,</span><span style="line-height:1.5;">如果只要求</span><span style="line-height:1.5;">F</span><span style="line-height:1.5;">（</span><span style="line-height:1.5;">n</span><span style="line-height:1.5;">）</span><span style="line-height:1.5;">,</span><span style="line-height:1.5;">是不是显得太水了，目测是的，所以我们现在来求一</span><span style="line-height:1.5;">求</span><span style="line-height:1.5;">SUM=[F</span><span style="line-height:1.5;">（</span><span style="line-height:1.5;">1</span><span style="line-height:1.5;">）</span><span style="line-height:1.5;">]^4+[F(2)]^4+....[F(n)]^4</span><span style="line-height:1.5;">的值，当然</span><span style="line-height:1.5;">1&lt;=n&lt;=10^9</span><span style="line-height:1.5;">，所以结果很大，要对</span><span style="line-height:1.5;">10^9+7</span><span style="line-height:1.5;">取余。雯神说“为什么不是求sigma(F[i]^我体重)呢？oh，it's too big”</span> </p> <p class="p0"> Ps<span style="line-height:1.5;">:[F(i)]^4</span><span style="line-height:1.5;">代表</span><span style="line-height:1.5;">F(i)</span><span style="line-height:1.5;">的</span><span style="line-height:1.5;">4</span><span style="line-height:1.5;">次方。</span> </p>

<p class="p0"> 输入一items数<span>n,</span><span><strong>当</strong></span><span><strong>n=0</strong></span><span><strong>时程序结束</strong></span> </p>

<p class="p0"> 输出<span>SUM%(10^9+7)</span><span>的值</span> </p>