AtCoder Beginner Contest 177

A - Don't be late

python

D, T, S = list(map(int, input().split()))

if T >= D / S:
    print('Yes')
else:
    print('No')

B - Substring

• 1000*1000なら全探索で間に合う

python

S = str(input())
T = str(input())

Ans=1000
for i in range(0, len(S)):
    Tmp=0
    if i + len(T) <= len(S):
      for j in range(0, len(T)):
          if S[i+j] != T[j]:
              Tmp += 1
      Ans = min(Ans, Tmp)
print(Ans)

C - Sum of product of pairs

• 2^100000 * 2^100000だと全探索では間に合わない
• O(N)にする
• グラフを作ると答えが見ます

C++

#include<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<utility>
#include<iomanip>
#include<map>
#include<queue>
#include<cmath>
#include<cstdio>

#define rep(i,n) for(int i=0; i<(n); ++i)
#define pai 3.1415926535897932384

using namespace std;
using ll =long long;
using P = pair<int,int>;

#define MOD 1000000007;

int main(int argc, const char * argv[]) {

    int N;
    ll A[200000];
    ll AA[200000];

    cin >> N;
    for(int i=0; i<N; i++){
        cin >> A[i];
        if(i==0) AA[i] = A[i];
        else AA[i] = (A[i] + AA[i-1]) % MOD;
    }

    ll Ans=0;
    for(int i=0; i<N; i++){
        Ans += (AA[i] * A[i]) % MOD;
        Ans = (Ans - A[i] * A[i]) % MOD;
        if(Ans < 0) Ans += MOD;
    }

    cout << Ans << endl;

    return 0;
}