A - Digits Sequence Dividing
CodeForces - 1107A
题意：一串字符分成两串，第一串比第二串小
题解：长度大于2和等于2考虑
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> #include<queue> #include<deque> #include<stack> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 2e5 + 10; int main() { int t; scanf("%d",&t); while(t--) { int n; string s; scanf("%d",&n); cin>>s; if(n == 2) { if(s[0] >= s[1]) puts("NO"); else { puts("YES"); printf("2 "); printf("%c %c ",s[0],s[1]); } } else { puts("YES"); puts("2"); printf("%c ",s[0]); for(int i=1;i<s.size();i++) printf("%c",s[i]); cout<<endl; } } }
B - Digital root
CodeForces - 1107B
题意：求第K个数位和为X的数字是多少
题解：找规律
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> #include<queue> #include<deque> #include<stack> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 2e5 + 10; int main() { int t; scanf("%d",&t); while(t--) { ll k,x; scanf("%lld %lld",&k,&x); printf("%lld ",(k - 1) * 9 + x); } }
C - Brutality
CodeForces - 1107C
题意：给n个数和一个值k，有一个26个字符的键盘，每个键不能连续按超过k次，现在给你一个按键序列和对于每次按键可以得到的权值，问最后可以得到的总权值是多少（注意如果按了k个a后按一个b的话又可以再按k次a）
题解：相同字段提取出来，累加上前k大
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> #include<queue> #include<deque> #include<stack> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 2e5 + 10; int n,k; int a[maxn]; int b[maxn]; char s[maxn]; queue<int>q; bool cmp(int x,int y) { return x > y; } ll sum = 0; int main() { scanf("%d %d",&n,&k); for(int i=0;i<n;++i) scanf("%d",&a[i]); scanf("%s",s); int tot = 0; for(int i=0;i<n;) { char tmp = s[i]; while(s[i] == tmp) { tot++; i++; } q.push(tot); tot = 0; } int id = 0; while(q.size()) { //printf("%d** ",q.front()); int ans = q.front(); int m = ans; q.pop(); int pos = 0; while(m--) { b[pos++] = a[id++]; } sort(b,b+ans,cmp); // for(int i=0;i<ans;i++) // printf("%d ",b[i]); // cout<<endl; if(k >= ans) { for(int i=0;i<ans;i++) sum += b[i]; } else { for(int i=0;i<k;i++) sum += b[i]; } } printf("%lld ",sum); }
D - Compression
CodeForces - 1107D
题意：十六进制的矩阵转换为二进制的矩阵，之后用x压缩矩阵，要求x整除n，并且原来矩阵中对应压缩后的矩阵的地方必须相等，求x
题解：矩阵转换之后按照每一行每一列的连续的1或者0分开，开始gcd，最后gcd的结果就是所求的x
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> #include<queue> #include<deque> #include<stack> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 5210; int a,b,c,d; void change(char ch) { int tmp; if('0' <= ch && ch <='9') { tmp = ch - '0'; } else { tmp = 10 + ch - 'A'; } if(tmp >= 8) { a = 1; tmp -= 8; } else a = 0; if(tmp >= 4) { b = 1; tmp -= 4; } else b = 0; if(tmp >= 2) { c = 1; tmp -= 2; } else c = 0; if(tmp >= 1) { d = 1; tmp -= 1; } else d = 0; } int gcd(int a,int b) { return b == 0 ? a : gcd(b,a % b); } char s[maxn][maxn]; char str[maxn]; queue<int>que; int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%s",str); int pos = 0; for(int len=0;len<n/4;len++) { change(str[len]); s[i][pos++] = a; s[i][pos++] = b; s[i][pos++] = c; s[i][pos++] = d; } } // for(int i=0;i<n;i++) // { // for(int j=0;j<n;j++) // printf("%d",s[i][j]); // cout<<endl; // } for(int i=0;i<n;i++) { int tmp = s[i][0]; int tot = 0; for(int j=0;j<n;j++) { if(tmp == s[i][j]) tot++; else { tmp = s[i][j]; que.push(tot); tot = 1; } } que.push(tot); } for(int j=0;j<n;j++) { int tmp = s[0][j]; int tot = 0; for(int i=0;i<n;i++) { if(tmp == s[i][j]) tot++; else { tmp = s[i][j]; que.push(tot); tot = 1; } } que.push(tot); } int minn = que.front(); que.pop(); while(!que.empty()) { minn = gcd(minn,que.front()); que.pop(); } printf("%d ",minn); }
E - Vasya and Binary String
CodeForces - 1107E
题意：一个长为n的01串，每次可以消去长度为len的连续相同字符，收益为alen,求消去整个串的收益最大值
题解：记忆化搜索可以更好的体现转移方程
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<vector> #include<string.h> using namespace std; #define ll long long #define mem(a, b) memset(a, b, sizeof(a)) const int maxn = 1e2 + 10; const int INF = 0x3f3f3f3f; ll dp[maxn][maxn][maxn]; string s; int a[maxn]; ll dfs(int l,int r,int k) { if(l > r) return 0; if(l == r) return a[k + 1]; if(dp[l][r][k] > 0) return dp[l][r][k]; dp[l][r][k] = dfs(l,r - 1,0) + a[k + 1]; for(int i = l; i < r; i++) { if(s[i] == s[r]) { dp[l][r][k] = max(dp[l][r][k],dfs(l,i,k + 1) + dfs(i + 1,r - 1,0)); } } return dp[l][r][k]; } int main() { int n; scanf("%d",&n); cin >> s; for(int i = 1; i <= n; i++) scanf("%d",&a[i]); memset(dp,-1,sizeof dp); //dfs(0,n - 1,0); cout << dfs(0,n - 1,0) <<endl; }