Fall hit Weekly Training 3

The difficulty of the topic added a bit. But it's still for the novice.

A.diff

The question was hard to see.

The fact is to classify the data on the line.

Let (A[i]-i) the number of equal to a class, any two of them meet a[i] = A[i-k]+k

A[i-flag] Indicates the number of a[i]-i = = I-flag.

The answer is N-max{a[i]}

1#include <cstdio>2#include <iostream>3#include <string.h>4 using namespacestd;5 6 #defineFlag 10000007 8 inta[2000010];9 intb[100010];Ten  One intMainvoid) A { -   intN; -   intMax; thescanf"%d", &n); -    while(N >0){ -     //printf ("****%d\n", n); -Memset (A,0,sizeof(a)); +      for(inti =0; I < n; ++i) scanf ("%d", &b[i]); -      for(inti =0; I < n; ++i) a[b[i]-i+flag]++; +Max =0; A      for(inti =0; I <2000000; ++i) at       if(A[i] >max) -Max =A[i]; -printf"%d\n", N-max); -scanf"%d", &n); -   } -   return 0; in}

B.dogs

This is the logical problem of a bad street. The answer is K.

1#include <cstdio>2 3 intMainvoid)4 {5   intm, N;6    while(SCANF ("%d%d", &m, &n) = =2) printf ("%d\n", n);7   return 0;8}

C.the Position

This is a mistake "no more than" means.

Not less than a person in front, it is only possible in [A+1, N]

No more than B-individuals are in the back, limiting only possible in [N-b, N]

Take the intersection, the number of which is min{n-a, b+1}

1#include <cstdio>2 3 #defineMin (a) (a) > (b)? (b): (a)4 5 intMainvoid)6 {7   intA, B, C;8    while(SCANF ("%d%d%d", &a, &b, &c) = =3) printf ("%d\n", Min (a-b,c+1));9   return 0;Ten}

D.restoring Password

is to test the simple processing of strings.

1#include <cstdio>2#include <string>3#include <iostream>4 using namespacestd;5 6 intMainvoid)7 {8   stringSt;9   strings;Ten   stringa[Ten]; One    while(cin>>St) { A      for(inti =0; I <Ten; ++i) cin>>A[i]; -      for(inti =0; I < the; i + =Ten){ -s =St; theS.erase (i+Ten, --i); -       if(I >0) S.erase (0, i); -       //cout<<s<<endl; -        for(intj =0; J <Ten; ++j) +     if(s = =A[j]) { -cout<<J; +        Break; A     } at     } -cout<<Endl; -   } -   return 0; -}

The 1.st.erase (x, y) represents the deletion of y from the beginning of the first X

2.string seems to use cin,cout more comfortable, anyway scanf,printf compiler did not pass

E.rotating

This problem seems to have been done on cogs years ago.

is a math problem, to simulate on the line.

But once it was submitted, I really can't believe it.

1#include <cstdio>2#include <cmath>3 4 #definePI 3.14159265358985 6 DoubleDisintNintk) {7   returnsqrt2-2*cos (k*2*pi/n));8 }9 Ten intMainvoid) One { A   intN; -   intN, S; -   DoubleCir; the   Doubleans; -scanf"%d", &N); -    for(intz =0; Z < N; ++z) { -scanf"%d%d", &n, &s); +CIR =0; -      for(inti =1; I <= N; ++i) Cir + = DIS (n,i) * (2*PI)/N; +Ans = S/n *Cir; A      for(inti =1; I <= s% n; ++i) ans + = dis (n,i) * (2*PI)/N; atprintf"%.2f\n", ans); -   } -   return 0; -}

Fall hit Weekly Training 3