vijos 1002

过河
考虑到l太大而n给出的数量很少所以我们能进行优化

for(int i = 1; i <= n; i ++) {
        a[i] = ((a[i] - a[i - 1]) % 100) + a[i - 1];
        b[a[i]] = 1;
    }

---

#include <cstdio>
#include <algorithm>
#define MAXN 10000010

using namespace std;

int b[MAXN], a[MAXN], f[MAXN];
int l, s, t, n;
int ans;

int read() {
    int f = 1, k = 0;
    char c = getchar();
    while (c < '0' || c > '9') {
        if(c == '-') {
            f = -1;
        }
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        k = k * 10 + c - '0';
        c = getchar();
    }
    return f * k;
}

int main() {
    l = read(), s = read(), t = read(), n = read();
    for(int i = 1; i <= n; i ++) {
        a[i] = read();
    }
    ans = 0;
    if(s == t) {
        for(int i = 1; i <= n; i ++) {
            if(a[i] % s == 0) {
                ans ++;
            }
        }
        printf("%d", ans);
        return 0;
    }
    sort (a + 1, a + n + 1);
    for(int i = 1; i <= n; i ++) {
        a[i] = ((a[i] - a[i - 1]) % 100) + a[i - 1];
        b[a[i]] = 1;
    }
    l = ((l - a[n]) % 100) + a[n];
    for(int i = 1; i <= 2 * l; i ++) {
        f[i] = MAXN;
    }
    for(int i = s; i <= l + t; i ++) {
        for(int j = s; j <= t && i - j >= 0; j ++) {
            f[i] = min(f[i], f[i - j] + b[i]);
        }
    }
    ans = MAXN;
    for(int i = l; i <= l + t; i ++) {
        ans = min(ans, f[i]);
    }
    printf("%d", ans);
    return 0;
}