過河
考慮到l太大而n給出的數量很少所以我們能進行優化

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 10000010using 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; }

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