题解
感觉自己通过刷水题混LOJ刷题量非常成功
首先是二进制枚举位，判是否合法
要写两个solve不是很开心，\(A\)不为1的直接记录状态\(f[i][j]\)为能否到达前\(i\)个分成\(j\)段，转移\(n^3\)
\(A\)为1的相当于在一张拓扑图上求到\(N\)的最短路是否小与\(B\)，连边方式即为如果\(sum[j] - sum[k]\)是二分值的一个子集则\(k\)到\(j\)有边
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 505
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,A,B,l;
int64 y[2005],sum[2005];
int dp[2005];
bool f[105][105];
void Solve1() {
int64 ans = 0;
for(int i = l ; i >= 0 ; --i) {
int64 t = ans + (1LL << i) - 1;
for(int j = 1 ; j <= N ; ++j) {
dp[j] = N + 1;
for(int k = 0 ; k < j ; ++k) {
if(((sum[j] - sum[k]) & t) == (sum[j] - sum[k])) dp[j] = min(dp[j],dp[k] + 1);
}
}
if(dp[N] > B) {ans |= (1LL << i);}
}
out(ans);enter;
}
void Solve2() {
int64 ans = 0;
for(int i = l ; i >= 0 ; --i) {
int64 t = ans + (1LL << i) - 1;
memset(f,0,sizeof(f));
f[0][0] = 1;
for(int j = 1 ; j <= N ; ++j) {
for(int k = 0 ; k < j ; ++k) {
if(((sum[j] - sum[k]) & t) == (sum[j] - sum[k])) {
for(int h = 1 ; h <= j ; ++h) {
f[j][h] |= f[k][h - 1];
}
}
}
}
bool flag = 0;
for(int k = A ; k <= B ; ++k) {
if(f[N][k]) {flag = 1;break;}
}
if(!flag) ans |= (1LL << i);
}
out(ans);enter;
}
void Init() {
read(N);read(A);read(B);
for(int i = 1 ; i <= N ; ++i) {read(y[i]);sum[i] = sum[i - 1] + y[i];}
int64 t = sum[N];
l = 0;
while(t) {++l;t >>= 1;}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
if(A == 1) Solve1();
else Solve2();
}