AHU 6th 校赛总结帖

A.随便搞，我偷懒用了一个map。手贱给again加一个感叹号还WA了一发，o(╯□╰)o

#include <cstdio>
#include <map>
 
using namespace std;
 
map<int,int> mp;
 
int main() {
    int n;  scanf("%d",&n);
    int mm,ss,v;
    scanf("%d:%d",&mm,&ss);
    v = mm * 60 + ss;
    for(int i = 1;i <= n;i++) {
        scanf("%d:%d",&mm,&ss);
        int tmp = mm * 60 + ss;
        if(mp[tmp] == 0) mp[tmp] = i;
    }
    if(mp[v] == 0) puts("Try Again");
    else printf("%d
",mp[v]);
    return 0;
}

B.n的范围只有200，模拟即可

#include <cstdio>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdlib>
 
using namespace std;
 
int a[205],maxv,maxi;
 
void findmax(int n) {
    maxv = -1;
    for(int i = 2;i <= n;i++) {
        if(a[i] > maxv) {
            maxv = a[i];
            maxi = i;
        }
    }
}
 
int main() {
    int n; scanf("%d",&n);
    int ans = 0;
    for(int i = 1;i <= n;i++) scanf("%d",&a[i]);
    findmax(n);
    while(maxv >= a[1]) {
        a[maxi]--; a[1]++;
        findmax(n);
        ans++;
    }
    if(ans == 0) printf("Oldbear, The best!
");
    else printf("%d
",ans);
    //system("pause");
    return 0;
}

C.Wzy说有规律可找，窝没发现(╯□╰)老老实实写了一个暴力

首先，只要确定a[1]和b[1]遍可以确定矩阵c了，确定a[1]就可以确定b[1] 了，所以只要枚举a[1]就好了

但是这里的数据范围有10^9，但是仔细观察可以发现，因为全是xor运算，a[1]可以一位一位的来确定，每一位只要枚举0和1就行了。

所以我们只要从高位到低位枚举a[1]的每一位，然后算出b,a中所有元素的对应位的值，就可以直接推出c

比赛的时候写的匆忙比较挫=。=

#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
const int maxn = 1030;
int a[maxn],b[maxn],c[maxn][maxn];
    int t,n,m;
 
inline int getbit(int v,int b) {
    if(v & (1 << b)) return 1;
    else return 0;
}
 
inline void setbit(int &s,int v,int b) {
    if(v) s |= (1 << b);
    else s &= ~(1 << b);
}
 
bool ok(int ss,int v1,int v2) {
    //printf("%d
",ss);
    setbit(a[1],v1,ss);
    setbit(b[1],v2,ss);
    for(int i = 2;i <= n;i++) {
        int bitc = getbit(c[i][1],ss);
        setbit(a[i],(bitc ^ v2),ss);
    }
 
    for(int i = 2;i <= m;i++) {
        int bitc = getbit(c[1][i],ss);
        setbit(b[i],bitc ^ v1,ss);
    }
 
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j <= m;j++) {
            int bitc = getbit(c[i][j],ss);
            int bita = getbit(a[i],ss);
            int bitb = getbit(b[j],ss);
            if(bitc != (bita ^ bitb)) return false;
        }
    }
    return true;
}
 
bool allok() {
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j <= m;j++) {
            //printf("%d %d %d %d
",c[i][j],a[i],b[j],a[i] ^ b[j]);
            if(c[i][j] != (a[i] ^ b[j])) return false;
        }
    }
    return true;
}
 
int main() {
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n;i++) {
            for(int j = 1;j <= m;j++) {
                scanf("%d",&c[i][j]);
            }
        }
        int ss;
        for(int ss = 30;ss >= 0;ss--) {
            if(getbit(c[1][1],ss) == 0) {
                if(ok(ss,0,0) || ok(ss,1,1)) continue;
            } else {
                if(ok(ss,0,1) || ok(ss,1,0)) continue;
            }
        }
        if(allok()) {
            for(int i = 1;i <= n;i++) {
                printf("%d",a[i]);
                if(i == n) putchar('
');
                else putchar(' ');
            }
            for(int i = 1;i <= m;i++) {
                printf("%d",b[i]);
                if(i == m) putchar('
');
                else putchar(' ');
            }
        } else {
            puts("I bet Tyrion made a mistake.");
        }
    }
    return 0;
}

D.动态规划，f[i]表示以i结尾的时候最大的价值。

f[i] =  f[j] + 1 >>> str[i] !=str[j]

　　　　f[j - 1] + sq(cc) >> str[i] == str[j] cc表示i到j之间str[j]的数量，相当于把这些相同的str[j]直接的元素全部去掉

　　　　j from i - 1 to 1

#include <cstdio>
#include <cstring>
#include <algorithm>
 
 
using namespace std;
 
const int maxn = 600;
char str[maxn];
int f[maxn];
 
inline int sq(int x) {
    return x * x;
}
 
int main() {
    int T; scanf("%d",&T);
    while(T--) {
        int ans = 1;
        scanf("%s",str + 1);
        int len = strlen(str + 1);
        memset(f,0,sizeof(f));
        f[1] = 1;
        for(int i = 2;i <= len;i++) {
            int maxv = 1;
            int cc = 1;
            for(int j = i - 1;j >= 1;j--) {
                if(str[i] != str[j]) {
                    maxv = max(maxv,f[j] + 1);
                } else {
                    cc++;
                    maxv = max(maxv,f[j - 1] + sq(cc));
                }
            }
            f[i] = maxv;
            ans = max(ans,f[i]);
        }
        printf("%d
",ans);
    }
    return 0;
}

E. 其实不是很难，就是利用前缀和和模运算有分配率来搞，比赛的时候没想到，最后安大的节操没有保住也有我的一部分责任o(╯□╰)o

设Si为序列前i项的和有sum(i,j) = Sj - Si-1

sum(i,j) % n = 0 >>> (Sj-Si-1)%n = 0 >> Sj % n == Si-1 %n

就是先扫一遍求S数组，然后扫一遍取摸，然后找L最小的两个相等值的位置啦=。=

#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
const int maxn = 100005;
int pos[maxn],a[maxn],sum[maxn];
 
int main() {
    int n; scanf("%d",&n);
    memset(pos,-1,sizeof(pos));
    int ansl = 2100000000,ansr,minlen = 2100000000;
    for(int i = 1;i <= n;i++) {
        scanf("%d",&a[i]);
        sum[i] = sum[i - 1] + a[i];
    }
    for(int i = 0;i <= n;i++) {
        sum[i] %= n;
        if(pos[sum[i]] == -1) {
            pos[sum[i]] = i;
        } else {
            if(pos[sum[i]] + 1 < ansl) {
                ansl = pos[sum[i]] + 1;
                ansr = i;
            }
        }
    }
    printf("%d %d
",ansl,ansr);
    return 0;
}

F.其实我不会做，目测Bo哥时限手一抖多打了一个0让我7s水过

完全暴力乱搞o(╯□╰)o

#include <cstdio>
#include <cstring>
#include <map>
#include <string>
#include <iostream>
#include <vector>
#include <algorithm>
 
using namespace std;

vector<string> dict;
vector<string> od;
int cc[100005];
 
int main() {
    ios::sync_with_stdio(false);
    int n; cin >> n;
    for(int i = 1;i <= n;i++) {
        string tmp;
        cin >> tmp;
        dict.push_back(tmp);
        od.push_back(tmp);
    }
    sort(dict.begin(),dict.end());
    for(int i = 0;i < od.size();i++) {
        int k = lower_bound(dict.begin(),dict.end(),od[i]) - dict.begin();
        int pos = k,maxv = cc[k],len = od[i].size(),maxi = k;
        while(pos < dict.size() && strncmp(od[i].c_str(),dict[pos].c_str(),len) == 0) {
            if(cc[pos] > maxv) {
                maxv = cc[pos]; maxi = pos;
            }
            pos++;
        }
        if(maxv == 0) {
            cout << od[i] << endl;
        } else cout << dict[maxi] << endl;
        cc[k]++;
    }
    //system("pause");
    return 0;
}

G.状态压缩动态规划，状态为当前的序列a，因为相同的数字没有意义，所以只要保留一个，因为有性质MAX{a1, a2, a3, ..., an}(表示取a1~an中值最大的一个)与LCM{a1, a2, a3, ..., an}(表示能被a1~an整除的最小正整数)相等，所以项数不会太多，我粗略的算了一下最多只有18。然后直接爆搜加记忆化就好。（这题赛后最后能搞定要多谢爱酱的指导╮(╯▽╰)╭，我一开始状态想错，后来又各种细节处理不好，编码能力和DP还是弱啊=。=）

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
 
#define WIN 1
#define LOSE 0
 
using namespace std;
 
const int maxn = 205;
int a[maxn];
 
//生成因数表
int factor[maxn],fcount;
void factor_maker(int n) {
    fcount = 0;
    for(int i = 2;i <= n;i++) {
        if(n % i == 0) factor[fcount++] = i;
    }
}
 
inline void setbit(int &n,int v,int b) {
    if(v) n |= (1 << b);
    else n &= ~(1 << b);
}
 
inline int getbit(int n,int b) {
    if(n & (1 << b)) return 1;
    else return 0;
}
 
inline int query(int n) {
    for(int i = 0;i < fcount;i++) if(factor[i] == n) return i;
}
 
int note[1 << 21];
 
int dfs(int state) {    //当前取的情况是state的时候先手是否可以赢
    int ret = WIN;
    if(~note[state]) return note[state];
    if(state == 0) return LOSE;
    for(int i = 0;i < fcount;i++) {
        int nstate = state;
        for(int j = 0;j < fcount;j++) {
            if(getbit(state,j) && factor[j] % factor[i] == 0) {
                setbit(nstate,0,j);
                if(factor[j] != factor[i]) {
                    setbit(nstate,1,query(factor[j] / factor[i]));
                } 
            }
        }
        if(nstate != state) {
            ret = dfs(nstate);
            if(ret == LOSE) break;
        }
    }
    return (note[state] = (!ret));
}
 
int main() {
    memset(note,-1,sizeof(note));
    int maxv = 0,n;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++) {
        scanf("%d",&a[i]);
        maxv = max(maxv,a[i]);
    }
    factor_maker(maxv);
    int ori = 0;
    for(int i = 1;i <= n;i++) {
        setbit(ori,1,query(a[i]));
    }
    int ret = dfs(ori);
    if(ret == WIN) puts("Poor Tyrion.");
    else puts("The Lannister always pays his debts.");
    return 0;
}

H.水题，直接来或者打表随便搞

#include <cstdio>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdlib>
 
using namespace std;
 
int val[300] = {
    1,3,7,13,19,27,39,49,63,79,87,91,103,109,123,133,139,147,169,181,183,187,207,219
,223,229,241,253,259,279,289,301,303,307,313,331,349,361,363,373,387,391,399,403
,423,439,447,459,463,469,481,499,507,511,523,529,543,553,567,571,583,589,601,613
,627,639,643,649,667,673,679,687,709,723,727,733,739,751,759,763,783,799,807,811
,819,841,843,853,859,867,889,901,903,907,927,931,943,949,963,973,979,
};
 
int main() {
    int n,t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n); printf("%d
",val[n - 1]);
    }
    //system("pause");
    return 0;
}