Hdu 5170 5171

Hdu 5170 5171

Hdu 5170 meaning: given four integers a, B, c, d. Compare the sizes of a ^ B and c ^ d,
If the data size is small, you can simply do it. Now the data is relatively large and can be compared with the logarithm on both sides;
But pay attention to accuracy issues!

#include 
  
   #include 
   
    #include 
    
     #include 
     
      #include #include 
      
       using namespace std;int a,b,c,d;int main(){ while(cin>>a>>b>>c>>d) { double tmp1=b*log10(a*1.0); double tmp2=d*log10(c*1.0); if(tmp1-tmp2>1e-9) printf(">\n"); else if(tmp2-tmp1>1e-9) printf("<\n"); else printf("=\n"); } return 0;}
      
     
    
   
  

Hdu 5171: given a set, the number in it may be repeated. Now we can retrieve two numbers a and B each time, and then add a + B to the set, this operation can be repeated k times
The final set is expressed as A = {A1, A2, A3 ..... An };
Sum = A1 + A2 + A2 + A3 + ...... + The maximum value of;
Greedy processing, that is, the maximum and secondary values in the original set are obtained each time. The data can be directly simulated when the data is small. Now the data is a little big. recursive formula:
A1 and b1 are used to represent the maximum values in the original set, and the secondary values

Add a1 + b1 to the set in the first operation
Add 2a1 + b1 to the set in the second operation
The third operation 3a1 + 2b1 joins the set.
Add 5a1 + 3b1 to the set for the fourth time.
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It is found to be a fib series, and the sum of the first k items can be obtained through matrix acceleration;

# Include
  
   
# Include
   
    
# Include
    
     
# Include
     
      
# Include const int MOD = 1e7 + 7; using namespace std; typedef long ll; int a [100100]; struct matrix {ll f [5] [5];}; matrix Co; // matrix mul (matrix a, matrix B) {matrix c; memset (c. f, 0, sizeof (c. f); for (int I = 0; I <4; I ++) for (int j = 0; j <4; j ++) for (int k = 0; k <4; k ++) {c. f [I] [j] + = (. f [I] [k] * B. f [k] [j]) % MOD; c. f [I] [j] % = MOD;} return c;} matrix quick_mod (matrix a, int B) {matrix s; memset (s. f, 0, sizeof (s. f); for (int I = 0; I <4; I ++) s. f [I] [I] = 1; while (B) {if (B & 1) s = mul (s, a); B> = 1; a = mul (a, a);} return s;} int main () {int n, k; Co. f [0] [0] = 1, Co. f [0] [1] = 1, Co. f [0] [2] = 0; Co. f [1] [0] = 1, Co. f [1] [1] = 0, Co. f [1] [2] = 0; Co. f [2] [0] = 1, Co. f [2] [1] = 1, Co. f [2] [2] = 1; while (cin> n> k) {ll ans = 0; int a1, b1; for (int I = 1; I <= n; I ++) {scanf ("% d", & a [I]); ans = (ans + a [I]) % MOD ;} sort (a + 1, a + 1 + n); a1 = a [n], b1 = a [n-1]; // maximum matrix tmp = Co; tmp = quick_mod (tmp, k); ll left = tmp. f [2] [0] * 1% MOD + tmp. f [2] [1] * 0% MOD + tmp. f [2] [2] * 0% MOD; ans = (ans + left * a1 % MOD) % MOD; tmp = Co; tmp = quick_mod (tmp, k-1 ); ll right = tmp. f [2] [0] * 1% MOD + tmp. f [2] [1] * 0% MOD + tmp. f [2] [2] * 0% MOD; ans = (ans + (right + 1) * b1 % MOD) % MOD; printf ("% I64d \ n ", ans);} return 0 ;}