Encryption (hard) CodeForces - 958C3 (树状数组)
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大意: 给定序列$a$, 要求将$a$分成$k$个非空区间, 使得区间和模$p$的和最小, 要求输出最小值.
$k$和$p$比较小, 直接暴力$dp$, 时间复杂度是$O(nklogp)$, 空间是$O(nk+kp)$
$dp[i][j]=min(...,f[j-1][s[i]-1]+1,f[j][s[i]],f[j][s[i]+1]-1+p,...)$
看了其他提交, 好像有$O(nk)$的做法.
#include <iostream> #include <sstream> #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 = 1e9+7, 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 #ifdef ONLINE_JUDGE const int N = 5e5+10; #else const int N = 111; #endif int n, k, p, a[N], s[N]; int dp[N][102]; struct BIT { int c[102]; BIT () {memset(c,0x3f,sizeof c);} void add1(int x, int v) { for (++x; x<=p; x+=x&-x) c[x]=min(c[x],v); } void add2(int x, int v) { for (++x; x; x^=x&-x) c[x]=min(c[x],v); } int qry1(int x) { int r=INF; for (++x; x; x^=x&-x) r=min(r,c[x]); return r; } int qry2(int x) { int r=INF; for (++x; x<=p; x+=x&-x) r=min(r,c[x]); return r; } } f1[102], f2[102]; int main() { scanf("%d%d%d", &n, &k, &p); REP(i,1,n) { scanf("%d", a+i); s[i]=(s[i-1]+a[i])%p; } f1[0].add1(0,0); f2[0].add2(0,0); REP(i,1,n) { REP(j,1,min(i,k)) { dp[i][j] = min(f1[j-1].qry1(s[i])+s[i],f2[j-1].qry2(s[i])+s[i]+p); } REP(j,1,min(i,k)) if (dp[i][j]<=INF) { f1[j].add1(s[i],dp[i][j]-s[i]); f2[j].add2(s[i],dp[i][j]-s[i]); } } printf("%d\n", dp[n][k]); }
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