第十九周 1.3-1.9
1.3
补个BC。
HDU 5608 function
题解这么说。复杂度不会。
1 #include <iostream> 2 #include <cstdio> 3 #include <map> 4 using namespace std; 5 typedef long long LL; 6 const LL mod = 1e9 + 7; 7 const LL maxn = 1e6 + 10; 8 map<LL,LL> M; 9 LL t[maxn]; 10 11 LL S1(LL n) 12 { 13 LL ret = n * ( n + 1 ) % mod * ( 2 * n % mod + 1 ) % mod * 166666668LL % mod; 14 ret = ( ret - 3 * n % mod * ( n + 1 ) % mod * 500000004LL % mod + mod ) % mod; 15 ret = ( ret + 2 * n ) % mod; 16 return ret; 17 } 18 19 LL S2(LL n) 20 { 21 if(n < maxn) return t[n]; 22 if(M.find(n) != M.end()) return M[n]; 23 LL ret = S1(n); 24 for(LL i = 2; i <= n; i++) 25 { 26 LL r = n / ( n / i ); 27 ret = (ret - S2(n/i) * ( r - i + 1 ) % mod + mod) % mod; 28 i = r; 29 } 30 return M[n] = ret; 31 } 32 33 void pre() 34 { 35 for(LL i = 1; i < maxn; i++) t[i] = (i - 1) * (i - 2) % mod; 36 for(int i = 1; i < maxn; i++) 37 for(int j = 2; i * j < maxn; j++) 38 t[i*j] = (t[i*j] - t[i] + mod) % mod; 39 for(int i = 1; i < maxn; i++) t[i] = ( t[i] + t[i-1] ) % mod; 40 return; 41 } 42 43 int main(void) 44 { 45 pre(); 46 int T; 47 scanf("%d", &T); 48 while(T--) 49 { 50 LL n; 51 scanf("%I64d", &n); 52 printf("%I64d\n", S2(n)); 53 } 54 return 0; 55 }
1.4-1.8
什么都没干。
1.9
怎么没人写C吖。
CF 615 C Running Track
n2预处理后缀。一路dp。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int suf[2][2222][2222], dp[2222]; 6 char s[2][2222], t[2222]; 7 int l[2222], r[2222]; 8 int len[2222]; 9 10 void ans_print(int pos) 11 { 12 if(pos <= 0) return; 13 ans_print(pos-len[pos]); 14 printf("%d %d\n", l[pos], r[pos]); 15 return; 16 } 17 18 int main(void) 19 { 20 scanf("%s%s", s[0] + 1, t + 1); 21 22 int ls = strlen(s[0]+1), lt = strlen(t+1); 23 for(int i = 1; i <= ls; i++) s[1][i] = s[0][ls-i+1]; 24 for(int i = 1; i <= ls; i++) 25 for(int j = 1; j <= lt; j++) 26 suf[0][i][j] = s[0][i] == t[j] ? suf[0][i-1][j-1] + 1 : 0 , 27 suf[1][i][j] = s[1][i] == t[j] ? suf[1][i-1][j-1] + 1 : 0 ; 28 29 for(int j = 1; j <= lt; j++) 30 { 31 dp[j] = 2222; 32 for(int i = 1; i <= ls; i++) 33 { 34 if(dp[j] > dp[j-suf[0][i][j]] + 1) 35 { 36 dp[j] = dp[j-suf[0][i][j]] + 1; 37 len[j] = suf[0][i][j]; 38 l[j] = i - len[j] + 1, r[j] = i; 39 } 40 if(dp[j] > dp[j-suf[1][i][j]] + 1) 41 { 42 dp[j] = dp[j-suf[1][i][j]] + 1; 43 len[j] = suf[1][i][j]; 44 l[j] = ls - i + len[j], r[j] = ls - i + 1; 45 } 46 } 47 } 48 49 if(dp[lt] == 2222) puts("-1"); 50 else 51 { 52 printf("%d\n", dp[lt]); 53 ans_print(lt); 54 } 55 56 return 0; 57 }
