J - Jesus Is Here HDU - 5459 (递推)

大意: 定义$f_1="c",f_2="ff",f_n=f_{n-2}+f_{n-1}$, 求所有"cff"的间距和.

记录c的个数, 总长度, 所有c到左边界距离和, 右边界距离和, 所有c的间距.

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 530600414, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 1e6+10;
int n, m, a[N];
char s[N];
int ans[N], num[N], len[N], L[N], R[N];

void init(int n) {
    num[1]=1,len[1]=1,len[2]=2;
    REP(i,3,n) {
        len[i]=(len[i-2]+len[i-1])%P;
        num[i]=(num[i-2]+num[i-1])%P;
        L[i]=(L[i-2]+L[i-1]+(ll)num[i-1]*len[i-2]%P)%P;
        R[i]=((ll)num[i-2]*len[i-1]%P+R[i-1]+R[i-2])%P;
        ans[i]=(ans[i-2]+ans[i-1]+(ll)(num[i-2]+R[i-2])*num[i-1]%P+(ll)L[i-1]*num[i-2]%P)%P;
    }
}

int main() {
    init(203333);
    int t;
    scanf("%d", &t);
    REP(i,1,t) {
        scanf("%d", &n);
        printf("Case #%d: %d\n", i,ans[n]);
    }
}