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  • 一个合法的三角形的充要条件是a<b+c,其中a为最长的一边,可以考虑找出所有不...增加两个假想长度为零的单位,所以方案数是C(tl+2,2), 假设a是增加长度以后最长的一边,r是tl减去给a增加长度之后剩下的长度, ...

    一个合法的三角形的充要条件是a<b+c,其中a为最长的一边,可以考虑找出所有不满足的情况然后用总方案减去不合法的情况。

    对于一个给定的总长度tl(一定要分完,因为是枚举tl,不分配的长度已经考虑过了),分成三份,因为可以有两份为零,采用插空法的时候

    增加两个假想长度为零的单位,所以方案数是C(tl+2,2),

    假设a是增加长度以后最长的一边,r是tl减去给a增加长度之后剩下的长度,

    那么不满足的情况的条件就可以写成a+tl-r>=b+c+r,很容易转化为r<=(a-b-c+tl)/2,显然r<=tl,如果r<0,

    说明tl全部加到a上也不会超过b+c的和,方案数为0。对于剩下的r,分成三份lb,lc和没用上的,类似之前总长度的分配,方案数为C(r+2,2),

     

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    
    ll cal(int a,int b,int c,int tl)
    {
        int r = (a-b-c+tl)>>1;
        if(r<0) return 0;
        if(r>tl) r = tl;
        return (ll)(r+1)*(r+2)>>1;
    }
    
    int main()
    {
        int a,b,c,l; scanf("%d%d%d%d",&a,&b,&c,&l);
        ll ans = 0;
        for(int i = 0; i <= l; i++){
            ans += (ll)(i+1)*(i+2)>>1;
            ans -= cal(a,b,c,i);
            ans -= cal(b,a,c,i);
            ans -= cal(c,a,b,i);
        }
        printf("%I64d",ans);
        return 0;
    }

     

    转载于:https://www.cnblogs.com/jerryRey/p/4752244.html

    展开全文
  • )) with parameters (array(float64, 1d, C), array(float64, 1d, C)) Known signatures: * (array(float32, 1d, A), array(float32, 1d, A)) -> float32 * parameterized [1] During: resolving callee type: ...
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  • Funny bug and a segfault

    2020-12-06 12:08:44
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  • 也可以从F,R(或C)求C(或R).选择相应的功能,并正确填写相关参数(均需要大于0),点击<计算>即可. 9.差分二阶LPF 这个电路常用于差分输出的DAC的LPF电路(默认使用最佳Q值计算).使用方法请参考第6点. 10.多层电感 此...
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  • 也可以从F,R(或C)求C(或R).选择相应的功能,并正确填写相关参数(均需要大于0),点击<计算>即可. 11.差分二阶LPF 这个电路常用于差分输出的DAC的LPF电路(默认使用最佳Q值计算).使用方法请参考第6点. 12.多层电感 此...
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