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; }