湖南工业大学个人选拔赛第二场 解题报告

A.连续子串和

贪心题，枚举每一个数字作为结束点。保留前i位的前缀和sum[i]，对于第i为结束的合法序列，其值为sum[i]-sum[i-K]，sum[i]-sum[i-K-1]，...，sum[i]-sum[0]，那么我们只需要对每一个 i 保留sum[0]到sum[i-K]的最小值即可。

代码如下：

Problem A#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int INF = 0x7fffffff;int N, K;int f[1000005];int solve() {    int ret = INF+1;    int Min = INF;    for (int i = K; i <= N; ++i) {        Min = min(Min, f[i-K]);        ret = max(ret, f[i]-Min);    }    return ret;}int main() {//    freopen("data.in", "r", stdin);//    freopen("data.out", "w", stdout);    while (scanf("%d %d", &N, &K) != EOF) {        for (int i = 1; i <= N; ++i) {            scanf("%d", &f[i]);            f[i] += f[i-1];        }        printf("%d\n", solve());    }    return 0;    }

B.谎言

水题，直接做一个取余操作就可以了，也可以把每一位对应的位权先计算出来乘上系数再加起来。

代码如下：

Problem B#include <cstring>#include <cstdlib>#include <cstdio>#include <algorithm>#include <iostream>using namespace std;char num[1005];int mod;void solve() {    int t = 0;    int len = strlen(num);    for (int i = 0; i < len; ++i) {        t = t*10 + num[i]-'0';        t %= mod;    }    printf(t ? "%d\n" : "YES\n", t);}int main() {//    freopen("data.in", "r", stdin);//    freopen("data.out", "w", stdout);    while (scanf("%s %d", num, &mod) != EOF) {        solve();    }    return 0;    }

C.费马定理

数学题，由于已经告知费马定理的存在，那么可以证明最小的满足要求的x一定是p-1的因子。可以假设如果x不是p-1的因子的话，那么令p-1=k*x+r，那么有a^(k*x) * a^r % p = 1，显然a^(k*x)%p = 1，而由于a^(p-1) % p = 1，所以a^r % p = 1，而0 < r < x并且x是满足条件最小的值，因此假设不成立。

代码如下：

Problem D#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>using namespace std;typedef long long LL;LL MOD;int a, p;const int INF = 0x7fffffff;int _pow(LL a, int b) {    LL ret = 1;    while (b) {        if (b & 1) {            ret *= a;            ret    %= MOD;        }        a *= a;        a %= MOD;        b >>= 1;    }    return ret;}void solve() {    p -= 1;    int ret = INF;    int LIM = (int)sqrt(double(p));    for (int i = 1; i <= LIM; ++i) {        if (p % i == 0) {            if (_pow(a, i) == 1) {                ret = min(ret, i);                }            if (_pow(a, p/i) == 1) {                ret = min(ret, p/i);            }        }    }    printf("%d\n", ret);}int main() {//    freopen("data.in", "r", stdin);//    freopen("data.out", "w", stdout);    while (scanf("%d %d", &a, &p) != EOF) {        MOD = p;        solve();        }    return 0;    }

D.平面划分

推理题：

   

新加入的一条直线与前面的直线都相交能够得到最多的空间划分。考虑到绿色的线是第三根插入的线，那么标号为1,2,3的线就是新区域的边界。对于如图加入第二个V型线，新增5个区域。如果增加第三个椭圆，新增4个区域。最后推出对于直线f[i] = f[i-1] + i；对于V型线f[i] = f[i-1] + 4*i-3；对于椭圆f[i] = f[i] + 2*i-2。

代码如下：

Problem D#include <cstdlib>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <ctime>using namespace std;const int MaxN = 1000000;long long f1[MaxN+5];long long f2[MaxN+5];long long f3[MaxN+5];void pre() {    f1[1] = f2[1] = f3[1] = 2;    for (int i = 2; i <= MaxN; ++i) {        f1[i] = f1[i-1] + i;    }    for (int i = 2; i <= MaxN; ++i) {        f2[i] = f2[i-1] + 4*i-3;    }    for (int i = 2; i <= MaxN; ++i) {        f3[i] = f3[i-1] + 2*i-2;        }}int main() {//    clock_t sta = clock();//    freopen("data.in", "r", stdin);//    freopen("data.out", "w", stdout);    int N;    pre();    while (scanf("%d", &N) != EOF) {        printf("%lld %lld %lld\n", f1[N], f2[N], f3[N]);    }//    clock_t end = clock();//    printf("%f\n", 1.0*(end-sta)/1000);    return 0;}

E.连续子串和续

利用第一题的做法加上二分搜索。设一个比例参数p，然后就是解决一个ai-p*bi的序列，至少连续K个，sum{ ai-p*bi } >= 0的问题了，二分枚举这个参数即可。

代码如下：

Problem E#include <cstdlib>#include <cstring>#include <cstdio>#include <iostream>#include <algorithm>using namespace std;const int INF = 0x7fffffff;const int MaxN = 1000005;int N, K;int A[MaxN], B[MaxN];double f[MaxN];bool Ac(double p) {    double Min = INF;    f[0] = 0;    for (int i = 1; i <= N; ++i) {        f[i] = f[i-1] + A[i]-p*B[i];        }    // 得到seq序列，现在要求一个连续的长度大于K的序列，使得总和大于等于0    for (int i = K; i <= N; ++i) {        Min = min(Min, f[i-K]);        if (f[i] - Min >= 0) {            return true;        }    }    return false;}double bsearch(double l, double r) {    double mid, ret;    while (r - l > 1e-8) {        mid = (l + r) / 2;        if (Ac(mid)) {            ret = mid;            l = mid + 1e-8;        } else {            r = mid - 1e-8;            }    }    return ret;}int main() {//    freopen("data.in", "r", stdin); //    freopen("data.out", "w", stdout);    while (scanf("%d %d", &N, &K) != EOF) {        for (int i = 1; i <= N; ++i) {            scanf("%d", &A[i]);        }        for (int i = 1; i <= N; ++i) {            scanf("%d", &B[i]);            }        printf("%.4f\n", bsearch(0, 1000));    }        return 0;}

转至http://www.cnblogs.com/Lyush/archive/2013/04/12/3017240.html