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  • SRM 553 DIV2

    250.

    解三元一次方程组。

    View Code
     1 class PlatypusDuckAndBeaver
     2 {
     3     public:
     4     int minimumAnimals(int a, int b, int c)
     5     {
     6         int k = (2*b+4*c-a)/2;
     7         int x = b - k;
     8         int y = c - k;
     9         return x + y + k;
    10     }
    11 };

    500.

    二分答案。

    View Code
     1 typedef long long LL;
     2 class Suminator
     3 {
     4     public:
     5     LL work(vector<int>f,int want)
     6     {
     7         stack<LL>S;
     8         S.push(0);
     9         for(int i=0;i<f.size();i++)
    10         {
    11 
    12             if(f[i] == 0)
    13             {
    14                 LL tmp = 0;
    15                 for(int j =S.size(),k=0;j>0&&k<2;k++,j--){
    16                     tmp += (LL)S.top();
    17                     S.pop();
    18                 }
    19                 S.push(tmp);
    20             }
    21             else S.push((LL)f[i]);
    22         }
    23         return S.top() - (LL)want;
    24     }
    25     int findMissing(vector <int> f, int want)
    26     {
    27         int n = f.size();
    28         int x=0;
    29         for(int i=0;i<n;i++)
    30         {
    31             if(f[i] == -1)
    32             {
    33                 x = i;
    34                 break;
    35             }
    36         }
    37         f[x]=0;
    38         if(work(f,want) == 0 )return 0;
    39         int l = 1,r = 1000000000;
    40         while(l<=r)
    41         {
    42             int m = (l + r)>>1;
    43             f[x] = m;
    44             LL tmp = work(f,want);
    45             if(tmp == 0)return m;
    46             else if(tmp > 0)r = m - 1;
    47             else l = m + 1;
    48         }
    49         return -1;
    50     }
    51 };
    by Farmer
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  • 原文地址:https://www.cnblogs.com/fzf123/p/2653238.html
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