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\n",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!\n");    else printf("%d\n",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\n",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\n",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('\n');                else putchar(' ');            }            for(int i = 1;i <= m;i++) {                printf("%d",b[i]);                if(i == m) putchar('\n');                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\n",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\n",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\n",val[n - 1]);    }    //system("pause");    return 0;}