Bestcoder Round #33

A and B practices are the same as the official solution

c I use Backpack +map, first the task according to the earliest start of time to sort, and then go to backpack, dp[j] mean J time can get the biggest score, and then over.

Code:

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    , Ty 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;}

Bestcoder Round #33