2017 Multi-University Training Contest 1

1001

题意：求2的m次方减去一这个数的位数

思路：很显然只有100 - 1， 1000 - 1 这类数减去1之后位数会降，2的m次方位数不可能是0，所以对m * log10(2)向下取整即可

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <string>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9;const int qq = 2e5 + 10;const double eps = 1e-9;int dp[qq];int main(){LL n, Cas = 0;while(scanf("%lld", &n) != EOF) {printf("Case #%lld: ", ++Cas);double ans = log10(2.) / log10(10.);LL t = floor(n * ans);printf("%lld\n", t);}return 0;}

1002

题意：给出n个字符串，对于字符a～z，赋值0～25，求在26进制下的最大值

思路：num[i][j]记录第i位上字符j有多少个，然后化成26进制，相当于比较字符a～z的26进制数的大小，然后贪心即可。

但是注意这里字符不能有前导零但是可以有字符串0，比如一个数据

26

a

b

c

...

z

那么其中必定有一个字符会被赋值为0，也就成了字符串0，但是其他情况不允许出现前导零，所以这种情况需要特判一下，然后对最小那个字符并且不在前导情况下赋值0

其他情况就是正常情况了

调了比较久，代码写的很...>..<

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <string>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)    memset(a, b, sizeof a)#define REP(i, x, n)    for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 50;int n;LL num[qq][30], cp[qq][26];int maxn[30], ct[30], cpmaxn[30];char st[qq];bool cmp(int x, int y) {    if(maxn[x] == maxn[y]) {        for(int i = maxn[x]; i >= 0; --i) {            if(num[i][x] == num[i][y])    continue;            return num[i][x] < num[i][y];        }        }    return maxn[x] < maxn[y];}bool vis[30], isapp[30];int main(){    int Cas = 0;    while(scanf("%d", &n) != EOF) {        mst(num, 0), mst(maxn, 0), mst(vis, false), mst(isapp, false);        printf("Case #%d: ", ++Cas);        for(int i = 0; i < n; ++i) {            scanf("%s", st);            int len = strlen(st);            if(len > 1) vis[st[0] - 'a'] = true;            for(int j = len - 1, k = 0; j >= 0; --j, k++) {                num[k][st[j] - 'a']++;                isapp[st[j] - 'a'] = true;                maxn[st[j] - 'a'] = max(maxn[st[j] - 'a'], k);            }        }        for(int i = 0; i < 26; ++i) {            cpmaxn[i] = maxn[i];            for(int j = 0; j <= maxn[i]; ++j) {                cp[j][i] = num[j][i];            }        }        for(int i = 0; i < 26; ++i) {            LL t = 0;            for(int j = 0; j <= maxn[i]; ++j) {                LL tmp = (num[j][i] + t) / 26;                num[j][i] = (t + num[j][i]) % 26;                t = tmp;                if(tmp > 0) {                    maxn[i] = max(maxn[i], j + 1);                }            }        }        for(int i = 0; i < 26; ++i) {            ct[i] = i;        }        int cnt = 0;        for(int i = 0; i < 26; ++i) {            if(isapp[i])    cnt++;        }        sort(ct, ct + 26, cmp);        int id = -1;        if(cnt == 26) {            for(int i = 0; i < 26; ++i) {                if(vis[ct[i]] == false) {                    id = i;                    break;                }            }        }        LL tmp = 0;        LL k = 25;        for(int i = 25; i >= 0; --i) {            if(id == i) {                continue;            }            LL ans = 0;            for(int j = cpmaxn[ct[i]]; j >= 0; --j) {//                printf("%c %lld\n", ct[i] + 'a', cp[j][ct[i]]);                ans = (ans * 26LL) % MOD;                ans = (ans + ((cp[j][ct[i]] * k) % MOD + MOD)) % MOD;            }            k = k - 1;            tmp = ((tmp + ans) % MOD + MOD) % MOD;        }        printf("%lld\n", tmp);    }    return 0;}

1006

参考：传送门

关键要想到构成了这种循环节

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <string>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9;const int qq = 1e5 + 10;int n, m;int a[qq], b[qq];LL num1[qq], num2[qq];void Init() {mst(num1, 0);mst(num2, 0);}int main(){int Cas = 0;while(scanf("%d%d", &n, &m) != EOF) {Init();for(int i = 0; i < n; ++i) {scanf("%d", a + i);}for(int i = 0; i < m; ++i ){scanf("%d", b + i);}int cycle = 0;for(int i = 0; i < n; ++i) {LL tot = 0;int k = i;while(a[k] != -1) {tot++;int t = k;k = a[k];a[t] = -1;}if(tot)num1[++cycle] = tot;}//puts("111");for(int i = 0; i < m; ++i) {LL tot = 0;int k = i;while(b[k] != -1) {tot++;int t = k;k = b[k];b[t] = -1;}if(tot)num2[tot]++;}//puts("111");LL ans = 1;for(int i = 1; i <= cycle; ++i) {LL ansl = 0;for(int j = 1; j * j <= num1[i]; ++j) {if(num1[i] % j == 0) {if(j * j == num1[i]) {ansl += num2[j] * j;} else {ansl += num2[j] * j + num2[num1[i] / j] * (num1[i] / j);}}}ans = (ans * ansl) % MOD;}//puts("111");printf("Case #%d: %lld\n", ++Cas, ans);}return 0;}

1011

题意：n双袜子放在橱柜里数字标号1～n，每天在橱柜拿一双袜子并且是当前数字最小的，当天晚上会把袜子扔进楼子里，当楼子里的袜子积累到n - 1双的时候，主人公会洗并且在第二天的晚上放进橱柜里，问第k天用的是那双袜子

思路：很显然k小于等于n时，答案就是k，k大于n是也知道前n次的结果，那么在第n天你用n这个袜子，之后n - 1天你可以用1 ～ n-1编号的袜子，那么在n + n - 1天你用的是n - 1这双袜子，之后n - 1天你可以用 1 ～ n - 2， n这袜子，规律很显然。

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <utility>#include <queue>#include <stack>#include <string>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9;const int qq = 2e5 + 10;int main(){LL n, m;int Cas = 0;while(scanf("%lld%lld", &n, &m) != EOF) {printf("Case #%d: ", ++Cas);if(n >= m) {printf("%lld\n", m);} else {m = m - n;LL tmp = m / (n - 1);LL t = n % (n - 1);if(t == 0) {printf("%lld\n", tmp % 2 == 1 ? n - 1 : n);} else {printf("%lld\n", t);}} }return 0;}