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。

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
const LL maxn = 1e6 + 10;
map<LL,LL> M;
LL t[maxn];

LL S1(LL n)
{
    LL ret = n * ( n + 1 ) % mod * ( 2 * n % mod + 1 ) % mod * 166666668LL % mod;
    ret = ( ret - 3 * n % mod * ( n + 1 ) % mod * 500000004LL % mod + mod ) % mod;
    ret = ( ret + 2 * n ) % mod;
    return ret;
}

LL S2(LL n)
{
    if(n < maxn) return t[n];
    if(M.find(n) != M.end()) return M[n];
    LL ret = S1(n);
    for(LL i = 2; i <= n; i++)
    {
        LL r = n / ( n / i );
        ret = (ret - S2(n/i) * ( r - i + 1 ) % mod + mod) % mod;
        i = r;
    }
    return M[n] = ret;
}

void pre()
{
    for(LL i = 1; i < maxn; i++) t[i] = (i - 1) * (i - 2) % mod;
    for(int i = 1; i < maxn; i++)
        for(int j = 2; i * j < maxn; j++)
            t[i*j] = (t[i*j] - t[i] + mod) % mod;
    for(int i = 1; i < maxn; i++) t[i] = ( t[i] + t[i-1] ) % mod;
    return;
}

int main(void)
{
    pre();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        LL n;
        scanf("%I64d", &n);
        printf("%I64d
", S2(n));
    }
    return 0;
}

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。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int suf[2][2222][2222], dp[2222];
char s[2][2222], t[2222];
int l[2222], r[2222];
int len[2222];

void ans_print(int pos)
{
    if(pos <= 0) return;
    ans_print(pos-len[pos]);
    printf("%d %d
", l[pos], r[pos]);
    return;
}

int main(void)
{
    scanf("%s%s", s[0] + 1, t + 1);

    int ls = strlen(s[0]+1), lt = strlen(t+1);
    for(int i = 1; i <= ls; i++) s[1][i] = s[0][ls-i+1];
    for(int i = 1; i <= ls; i++)
        for(int j = 1; j <= lt; j++)
            suf[0][i][j] = s[0][i] == t[j] ? suf[0][i-1][j-1] + 1 : 0 ,
            suf[1][i][j] = s[1][i] == t[j] ? suf[1][i-1][j-1] + 1 : 0 ;

    for(int j = 1; j <= lt; j++)
    {
        dp[j] = 2222;
        for(int i = 1; i <= ls; i++)
        {
            if(dp[j] > dp[j-suf[0][i][j]] + 1)
            {
                dp[j] = dp[j-suf[0][i][j]] + 1;
                len[j] = suf[0][i][j];
                l[j] = i - len[j] + 1, r[j] = i;
            }
            if(dp[j] > dp[j-suf[1][i][j]] + 1)
            {
                dp[j] = dp[j-suf[1][i][j]] + 1;
                len[j] = suf[1][i][j];
                l[j] = ls - i + len[j], r[j] = ls - i + 1;
            }
        }
    }

    if(dp[lt] == 2222) puts("-1");
    else
    {
        printf("%d
", dp[lt]);
        ans_print(lt);
    }

    return 0;
}