hdu 2542 补兵(二分，模拟)

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题意：就是补兵。。

先用二分查找得到两端。然后压缩掉不符合的端点。。

注意，估计最大时间的时候，很可能会爆longlong（猜的），所以要用 (H-P)/rate, rate是每秒掉血。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <vector>#include <queue>using namespace std;#define REP(i,s,t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i,s,t) for(int (i)=(s);(i)>=(t);--(i))#define INF 0x7FFFFFFFconst int maxn = 100000;int max_harm[1005];int accum_harm[1005];int A, T, P, H;int max_interval, sum;bool check(int x) {    int m = 0;    for (int i=1;i<=max_interval;++i) {        if (accum_harm[i]) {            m += x/i*accum_harm[i];        }    }    if (m + P>= H) return true;    return false;}bool check2(int x) {    int m = 0;    for (int i=1;i<=max_interval;++i) {        if (accum_harm[i]) {            m += x/i*accum_harm[i];        }    }    if (m >= H) return true;    return false;}int go1(int l, int r) {    int mid;    while (l<r) {        mid = l + (r-l)/2;        if (check(mid))            r = mid;        else            l = mid+1;    }    if(check(r))        return r;    return -1;}int go2(int l, int r) {    int mid;    while (l<r) {        mid = l + (r-l)/2;        if (check2(mid))            r = mid;        else            l = mid+1;    }    if(check2(r))        return r;    return -1;}int main() {    //freopen("input.in", "r", stdin);    int n;    while (scanf("%d",&n) && n) {        memset(accum_harm, 0, sizeof(accum_harm));        memset(max_harm, 0, sizeof(max_harm));        max_interval = 100;        double rate = 0;        REP(i, 1, n-1) {            scanf("%d%d",&A, &T);            max_interval = max(max_interval, T);            max_harm[T] = max(max_harm[T], A);            accum_harm[T] += A;            rate += A*1.0/T;        }        scanf("%d%d",&P,&H);        int low = 1, up = H/rate+100;        int g1 = go1(low, up);        int g2 = go2(low, up);        int k1 = -1, k2 = -1;        //cout << k1 << ' ' << k2 << endl;        if (g1 && g2) {            REP(i, g1, g2) {                int mx = 0;                for (int j=1;j<=max_interval;++j) {                    if (i%j == 0 && mx < max_harm[j])                        mx = max_harm[j];                }                if (mx > P) break;                if (check2(i-1) == false && check(i) == true && mx <= P) {                    k1 = i;                    break;                }            }            UREP(i, g2, g1) {                int mx = 0;                for (int j=1;j<=max_interval;++j) {                    if (i%j == 0 && mx < max_harm[j])                        mx = max_harm[j];                }                if (check2(i-1) == false && check(i) == true && mx <= P) {                    k2 = i;                    break;                }            }        }        if (k1 == -1 || k2 == -1) printf("Impossible\n");        else            printf("%d %d\n", k1, k2);    }    return 0;}

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