给出一个数 k, 问用 k 个斐波那契数相加, 得不到的数最小是几.
- #include <cstdio>
- #include <cstring>
- #include <cmath>
- #include <algorithm>
- #include <vector>
- #include <string>
- #include <map>
- #include <set>
- #include <cassert>
- #include<bits/stdc++.h>
- using namespace std;
- #define rep(i,a,n) for (int i=a;i<n;i++)
- #define per(i,a,n) for (int i=n-1;i>=a;i--)
- #define pb push_back
- #define mp make_pair
- #define all(x) (x).begin(),(x).end()
- #define fi first
- #define se second
- #define SZ(x) ((int)(x).size())
- typedef vector<int> VI;
- typedef long long ll;
- typedef pair<int,int> PII;
- const ll mod=998244353;
- ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
- // head
- int _,n;
- namespace linear_seq {
- const int N=10010;
- ll res[N],base[N],_c[N],_md[N];
- vector<int> Md;
- void mul(ll *a,ll *b,int k) {
- rep(i,0,k+k) _c[i]=0;
- rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
- for (int i=k+k-1;i>=k;i--) if (_c[i])
- rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
- rep(i,0,k) a[i]=_c[i];
- }
- int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
- // printf("%d\n",SZ(b));
- ll ans=0,pnt=0;
- int k=SZ(a);
- assert(SZ(a)==SZ(b));
- rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
- Md.clear();
- rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
- rep(i,0,k) res[i]=base[i]=0;
- res[0]=1;
- while ((1ll<<pnt)<=n) pnt++;
- for (int p=pnt;p>=0;p--) {
- mul(res,res,k);
- if ((n>>p)&1) {
- for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
- rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
- }
- }
- rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
- if (ans<0) ans+=mod;
- return ans;
- }
- VI BM(VI s) {
- VI C(1,1),B(1,1);
- int L=0,m=1,b=1;
- rep(n,0,SZ(s)) {
- ll d=0;
- rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
- if (d==0) ++m;
- else if (2*L<=n) {
- VI T=C;
- ll c=mod-d*powmod(b,mod-2)%mod;
- while (SZ(C)<SZ(B)+m) C.pb(0);
- rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
- L=n+1-L; B=T; b=d; m=1;
- } else {
- ll c=mod-d*powmod(b,mod-2)%mod;
- while (SZ(C)<SZ(B)+m) C.pb(0);
- rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
- ++m;
- }
- }
- return C;
- }
- int gao(VI a,ll n) {
- VI c=BM(a);
- c.erase(c.begin());
- rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
- return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
- }
- };
- int main() {
- while (~scanf("%d",&n)) {
- vector<int>v;
- v.push_back(4);
- v.push_back(12);
- v.push_back(33);
- v.push_back(88);
- v.push_back(232);
- v.push_back(609);
- v.push_back(1596);
- printf("%d\n",linear_seq::gao(v,n-1));
- }
- }
打表:
- #include <stdio.h>
- #include <string.h>
- using namespace std;
- typedef long long ll;
- int dp[200][2000];
- int main()
- {
- ll c[1100]={0,1,1};
- for(int i=2;i<1020;i++)
- c[i]=(c[i-1]+c[i-2]);
- memset(dp, 0, sizeof(dp));
- dp[0][0] = 1;
- for (int i = 0; i <= 40; i++) // 枚举总类
- {
- for (int num = 1; num <= 40; num++) // 枚举个数
- {
- for (int j = c[i]; j <= 1000; j++) // 枚举容量
- {
- dp[num][j] += dp[num - 1][j - c[i]];
- }
- }
- }
- for(int i=0;i<=40;i++)
- for(int j=1;j<=1000;j++)
- if(dp[i][j]==0)
- {
- printf("%d\n",j);break;
- }
- return 0;
- }
来源: http://www.bubuko.com/infodetail-2768963.html