A和B做法和官方題解一樣

C題我是用背包+map，先把任務按最早開始的時間進行排序，然后去背包，dp[j]表示j時間能得到最大的得分，然后就過了。。

代碼：

A：

#include <stdio.h> #include <string.h> #include <algorithm> using namespace std;int n, b;char str[205]; int ans[205];int get(char c) {if (c >= '0' && c <= '9') return c - '0';return c - 'a' + 10; }void print(int c) {if (c >= 0 && c <= 9) printf("%d", c);else printf("%c", c + 'a' - 10); }int main() {while (~scanf("%d%d", &n, &b)) {memset(ans, 0, sizeof(ans));while (n--) {scanf("%s", str);int len = strlen(str);for (int i = len - 1; i >= 0; i--) {ans[len - i - 1] += get(str[i]);}}for (int i = 0; i < 204; i++) {ans[i] %= b;}int i;for (i = 204; i >= 0; i--) {if (ans[i])break;}if (i == -1) i++;for (int j = i; j >= 0; j--)print(ans[j]);printf("\n");}return 0; }
B：

#include <cstdio> #include <cstring> #include <algorithm> using namespace std;typedef long long ll;ll n, p;ll muti(ll a, ll n, ll p) {ll r = 0;while(n){if(n&1){r += a;if(r >= p) r -= p;}n >>= 1;a += a;if(a >= p) a -= p;}return r; }ll pow_mod(ll x, ll k) {ll ans = 1;x %= p;while (k) {if (k&1) ans = muti(ans, x, p);x = muti(x, x, p);k >>= 1;}return ans; }int main() {while (~scanf("%I64d%I64d", &n, &p)) {if (n == 1) printf("%I64d\n", n % p);else printf("%I64d\n", ((pow_mod(2LL, n) - 2) % p + p) % p);}return 0; }
C：

#include <cstdio> #include <cstring> #include <map> #include <algorithm> using namespace std;typedef long long ll; const int N = 35;int n; ll w;struct Q {int t, v, l, s;void read() {scanf("%d%d%d", &t, &v, &l);s = max(0, l - t);} } q[N];map<int, ll> dp[2]; map<int, ll>::iterator it;bool cmp(Q a, Q b) {if (a.s == b.s) return a.t < b.t;return a.s < b.s; }int main() {while (~scanf("%d%I64d", &n, &w)) {ll sum = 0;for (int i = 0; i < n; i++) {q[i].read();sum += q[i].v;}if (sum < w) {printf("zhx is naive!\n");continue;}sort(q, q + n, cmp);int now = 0, pre = 1;dp[now].clear();dp[now][0] = 0;for (int i = 0; i < n; i++) {swap(now, pre);dp[now].clear();for (it = dp[pre].begin(); it != dp[pre].end(); it++) {int pt = it->first;ll pw = it->second;dp[now][pt] = max(dp[now][pt], pw);int ut = max(pt, q[i].s) + q[i].t;dp[now][ut] = max(dp[now][ut], pw + q[i].v);}}for (it = dp[now].begin(); it != dp[now].end(); it++) {if (it->second >= w) {int ans = it->first;printf("%d\n", ans);break;}}}return 0; }

總結

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