2017 Multi-University Training Contest

1001

题意：给出一个n， a， b，代表有n道题，Derek and Alfia的得分，然后出给Derek and Alfia作答序列，每道题只包含ABC三个答案，问是否存在一种答案序列使得Derek and Alfia的得分恰好等于a和b

思路：统计出Derek and Alfia统一道题答案相同的次数x，然后y是作答答案不相同的题目个数，然后YY一下

#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 ft first#define sd second#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 80000 + 10;char st[qq], mt[qq];int main(){int t;scanf("%d", &t);while(t--) {int n, a, b;scanf("%d%d%d", &n, &a, &b);scanf("%s%s", st, mt);int len = strlen(st);int cnt = 0;for(int i = 0; i < n; ++i) {if(st[i] == mt[i])cnt++;}int x = cnt, y = n - cnt;int tmp = min(x, min(a, b));a -= tmp, b -= tmp;if(y >= a + b) {puts("Not lying");} else {puts("Lying");}}return 0;}

1003

题意：给出序列a，和序列b，然后要求an + 1 到 a2n，从an + 1开始求一直求到a2n，对于每个ai，你可以在b数组中选一个数那么你可以这个ai可以小于等于aj - j (bk <= j < i)

求出最大和

思路：如果b数组是n的全排列，那么b数组其实是没有用的，但这题不是，b数组中的数值有重复的，那么实际上我们从大到小贪心即可，dp[i]代表从i位置开始一直到n位置中ai - i的最大值，对于每次更新的ai我们用一个maxn记录an + 1 到 i的最大值，那么每次贪心的时候就是maxn 和 dp[i]中选一个即可

#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 ft first#define sd second#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 250000 + 10;const int MOD = 1e9 + 7;LL dp[qq], num[qq], b[qq];LL maxn;int n;bool cmp(int x, int y) {if(dp[x] == dp[y])return 0;return dp[x] < dp[y];}int main(){while(scanf("%d", &n) != EOF) {for(int i = 1; i <= n; ++i) {scanf("%lld", num + i);num[i] = num[i] - i;}for(int i = 1; i <= n; ++i) {scanf("%lld", b + i);}dp[n + 1] = 0;for(int i = n; i >= 1; --i) {dp[i] = max(dp[i + 1], num[i]);}sort(b + 1, b + 1 + n, cmp);maxn = 0;LL ans = 0;LL cnt = n + 1;for(int i = n; i >= 1; --i, cnt++) {LL tmp = max(maxn, dp[b[i]]);ans = (ans + tmp) % MOD;maxn = max(maxn, tmp - cnt);}printf("%lld\n", ans % MOD);}return 0;}

1004

队友之前课设做过关于拼图的恰好有这个结论存在

就是求拼图序列的逆序数即可，为偶就可以拼出来，否则不可以

对于这里如何构造序列，其实这里不用构造...

只需要算出逆序数

#include<cstdio>#include<cstring>#include<cmath>#include<string>#include<iostream>#include<algorithm>#include<queue>#include<vector>#include<stack>using namespace std;typedef long long LL;int main() {    int t,n,m,p;    scanf("%d",&t);    while(t--) {        scanf("%d%d%d",&n,&m,&p);        int ans=0;        int cnt=n*m-1,cnt1=n*m-1,top=1;        while(1) {            if(top>cnt) break;            cnt1--;            top++;            int pp=1;            for(int i=1;;i++) {                pp=pp+p-1;                if(pp>cnt1) break;                ans=(ans+pp-1)%2;                cnt1--;                top++;            }        }        if(!ans) printf("YES\n");        else printf("NO\n");    }    return 0;}

1009

题意：给出一个a序列，然后你需要构造一个b序列，1 <= bi <= ai，要求b序列中任意区间的gcd都要 >= 2，问有多少种b序列满足条件

思路：之前做过一个求一个序列中gcd = 1的子序列有多少个的题

来自：传送门

仔细想想其实和这题有一点点类似，但是求解方法是不一样的、

比赛中没有想出来，最后参考了 传送门  这位聚聚的方法才知道做法、

关于莫比乌斯反演：传送门

其实最终的方法还是容斥，但关键就关键在如何容斥、这为聚聚的方法已经讲得很明白了，就不赘述了、

#include <cstdio>#include <cstring>#include <cmath>#include <sstream>#include <iostream>#include <algorithm>#include <string>#include <stack>#include <queue>#include <vector>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define ft first#define sd second#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 1e5 + 10;const int MOD = 1e9 + 7;const int INF = 1e9 + 10;bool isprime[qq];int num[qq], prime[qq], cnt = 0;int mu[qq];void Init() {for(int i = 2; i < qq; ++i) {if(!isprime[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && i * prime[j] < qq; ++j) {isprime[i * prime[j]] = true;if(i % prime[j]) {mu[i * prime[j]] = -mu[i];} else {mu[i * prime[j]] = 0;break;}}}}LL QuickPow(LL a, LL b) {LL ans = 1;while(b > 0) {if(b & 1)ans = (ans * a) % MOD;a = (a * a) % MOD;b >>= 1;}return ans;}int maxn, minx, n;LL Solve() {for(int i = 2; i <= maxn; ++i) {num[i] += num[i - 1];}LL ans = 0;for(int i = 2; i <= minx; ++i) if(mu[i]) {LL tmp = -mu[i];int m = maxn / i;for(int j = 1; j <= m; ++j) {int l = i * j, r = min(i * j + i - 1, maxn);tmp = (tmp * QuickPow(j, num[r] - num[l - 1])) % MOD;}ans = (ans + tmp) % MOD;}return (ans % MOD + MOD) % MOD;}int main(){Init();int T, Cas = 0;scanf("%d", &T);while(T--) {scanf("%d", &n);mst(num, 0);maxn = -INF, minx = INF;for(int i = 0; i < n; ++i) {int x;scanf("%d", &x);num[x]++;maxn = max(maxn, x), minx = min(minx, x);}printf("Case #%d: %lld\n", ++Cas, Solve());}return 0;}

1011

题意：给出n个整点的坐标，问可以构成多少个正多边形

思路：可知在平面上整点的正多边形只有正方形，然后枚举下就好了，注意考虑斜着的正方形

#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 ft first#define sd second#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 400 + 10;const int MOD = 1e9 + 7;bool vis[qq][qq];pill p[qq + 100];int n;int main(){while(scanf("%d", &n) != EOF) {mst(vis, false);for(int x, y, i = 0; i < n; ++i) {scanf("%d%d", &x, &y);vis[x + 100][y + 100] = true;p[i] = mk(x + 100, y + 100);}int cnt = 0;for(int l = 1; l <= 200; ++l) {for(int i = 0; i + l <= 200; ++i) {for(int j = 0; j + l <= 200; ++j) {if(vis[i][j] && vis[i + l][j] && vis[i][j + l] && vis[i + l][j + l]) {cnt++;}}}}sort(p, p + n);for(int i = 0; i < n; ++i) {for(int y = 1; p[i].ft + y <= 200; ++y) {for(int x = 1; p[i].sd + x <= 200; ++x) {int a = p[i].ft;int b = p[i].sd;if(a - x >= 0 && b + x + y <= 200 && vis[a - x][b + y] && vis[a + y][b + x] && vis[a - x + y][b + x + y]) {cnt++;}}}}printf("%d\n", cnt);}return 0;}