设计素材网站名称,搭建域名服务器,广州seo代理商,南水北调建设管理局网站atcoder_abc\AtCoder Beginner Contest 310\E_NAND_repeatedly
//题意#xff1a;给定一个n长度的01串#xff0c;计算f(l,r)(lr,l在1~n,r在1~n)的和,f的计算(ai,a(i1))运算,有0就为1,11为0
//若f(l,r)1,则f(l,r-1)为0或sr为0,即只取决于上一位的情况和当前位#xff…atcoder_abc\AtCoder Beginner Contest 310\E_NAND_repeatedly
//题意给定一个n长度的01串计算f(l,r)(lr,l在1~n,r在1~n)的和,f的计算(ai,a(i1))运算,有0就为1,11为0
//若f(l,r)1,则f(l,r-1)为0或sr为0,即只取决于上一位的情况和当前位枚举右端点对答案的贡献即有多少左端点使得f(i,r)1;
#includebits/stdc.h#includeiostream#includealgorithm#includemap#includeset#includequeue#includecstring#includemath.h#includemap#includevector#includestack#includeunordered_mapusing namespace std;#define endl \ntypedef pairint,int pr;#define int long long#define int128 __int128_t#define ll long long#define fr(i,l,r) for(int il;ir;i)#define fer(i,x) for(int ie.head[x];i;ie.next[i])#define ufr(i,n,z) for(int i n;i z; i--)#define pb(x) push_back(x)#define all(a) a.begin(),a.end()#define fi first#define se secondconst int N 1e610;const int mod998244353,infLONG_LONG_MAX;ll poww(ll x, ll k){if(k 1) return x;if(k 0) return 1;ll tmp poww(x, k / 2);if(k 1) return tmp * tmp % mod * x % mod;return tmp * tmp % mod;}templatesize_t sizestruct Road{int to[size], next[size], head[size], cnt 1;ll w[size];void add(int x, int y, ll ww){to[cnt] y;w[cnt] ww;next[cnt] head[x];head[x] cnt ;}void clear(int n){for(int i 0; i n; i ){head[i] 0;}cnt 1;}};templatesize_t sizestruct Combinatorial_number{int fact[size], finv[size]; // fact[x]是x的阶乘inv[x]是fact[x]的逆元void init(){fact[0] 1;for(int i 1; i (int)size; i ){fact[i] 1ll * fact[i - 1] * i % mod;}finv[size - 1] poww(fact[size - 1], mod - 2);for(int i size - 2; i 0; i --){finv[i] 1ll * finv[i 1] * (i 1) % mod;}}ll C(int n, int m){if(m n || n 0 || m 0) return 0;return 1ll * fact[m] * finv[n] % mod * finv[m - n] % mod;}ll A(int n, int m){if(m n || n 0 || m 0) return 0;return 1ll * fact[m] * finv[m - n] % mod;}ll llC(int n, int m){//求解不进行取余mod的可能爆longlong的C(n,m)值;if(m n || n 0 || m 0) return 0;swap(n, m);long long sum 1;if (m n - m) m n - m;n n - m 1;for (int i 1; i m; i ){sum * n ;sum / i;}return sum;}};templatesize_t sizestruct Prime{int con[size], tot 0, vis[size];void init(){for(int i 2; i (int)size; i ){if(vis[i] 0) con[ tot] i;for(int j 1; j tot i * con[j] (int)size; j ){vis[i * con[j]] con[j];if(i % con[j] 0) break;}}}};// Road N e; // 无向图 * 2// Combinatorial_number N comb;// Prime N prime;//题意给定一个n长度的01串计算f(l,r)(lr,l在1~n,r在1~n)的和,f的计算(ai,a(i1))运算,有0就为1,11为0//若f(l,r)1,则f(l,r-1)为0或sr为0,即只取决于上一位的情况和当前位枚举右端点对答案的贡献即有多少左端点使得f(i,r)1;//int n,m;int a[N];int dp[N][2]; //dp[i][0/1]表示f(l,i)0的数量或f(l,i)1的数量void solve(){cinn;string s;cins;s s;int ans0;fr(i,1,s.size()-1){if(s[i]0){dp[i][0]1; //本身dp[i][1]dp[i-1][0]dp[i-1][1]; //考虑上一位有多少f(l,i-1)0的数量或f(l,i-1)1的数量}else{dp[i][0]dp[i-1][1];dp[i][1]dp[i-1][0]1;}ansdp[i][1];}coutans\n;}signed main(){int t1;// cint;while(t--) solve();return 0;}
atcoder_abc\AtCoder Beginner Contest 313\D_Odd_or_Even
//题意交互题给定n可以选择k个数k为奇数且kn程序告诉k个数的异或值求可以唯一确定的n长度序列数值只会为0或1
//思路以n4,k3为例先查询1 2 3,1 3 4,1 2 4三个结果的异或值将三个结果异或就可以得知a1的值同理前k1个数也可推出
//要知道第k2个数查询3,4..k2的异或值再异或a3,a4...ak1即可
//k为奇数是为了保证在k次查询中除查询的其他均出现偶数次
#include bits/stdc.husing namespace std;using LL long long;int main(void) {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int n, k;cin n k;vectorint ans(n);auto solve [](int l, int r) {vectorint tmp(r - l);vectorint q;for (int i l; i r; i) {q.clear();for (int j l; j r; j)if (j ! i)q.push_back(j);cout ?;for (auto i : q)cout i 1;cout endl;cin tmp[i - l];}for (int i l; i r; i) {for (int j l; j r; j) {if (j ! i) {ans[i] ^ tmp[j];}}}};solve(0, k 1);for (int i k 1; i n; i) {cout ?;for (int j i; j i - k; --j) {cout j 1;}cout endl;cin ans[i];for (int j i - 1; j i - k; --j)ans[i] ^ ans[j];}cout !;for (auto i : ans)cout i;cout endl;return 0;}
//题意给定一个字符串由()?组成可以将?替换成(或)求能组成满足()的个数 //状态dp[i][j]表示在i位置,(记作1)表示-1j表示前缀和
#includebits/stdc.h#includeiostream#includealgorithm#includemap#includeset#includequeue#includecstring#includemath.h#includemap#includevector#includestack#includeunordered_mapusing namespace std;#define endl \ntypedef pairint,int pr;#define int long long#define int128 __int128_t#define ll long long#define fr(i,l,r) for(int il;ir;i)#define ufr(i,n,z) for(int i n;i z; i--)#define pb(x) push_back(x)#define all(a) a.begin(),a.end()#define fi first#define se secondconst int N 1e610;const int mod998244353,infLONG_LONG_MAX;ll poww(ll x, ll k){if(k 1) return x;if(k 0) return 1;ll tmp poww(x, k / 2);if(k 1) return tmp * tmp % mod * x % mod;return tmp * tmp % mod;}templatesize_t sizestruct Road{int to[size], next[size], head[size], cnt 1;ll w[size];void add(int x, int y, ll ww){to[cnt] y;w[cnt] ww;next[cnt] head[x];head[x] cnt ;}void clear(int n){for(int i 0; i n; i ){head[i] 0;}cnt 1;}};templatesize_t sizestruct Combinatorial_number{int fact[size], finv[size]; // fact[x]是x的阶乘inv[x]是fact[x]的逆元void init(){fact[0] 1;for(int i 1; i (int)size; i ){fact[i] 1ll * fact[i - 1] * i % mod;}finv[size - 1] poww(fact[size - 1], mod - 2);for(int i size - 2; i 0; i --){finv[i] 1ll * finv[i 1] * (i 1) % mod;}}ll C(int n, int m){if(m n || n 0 || m 0) return 0;return 1ll * fact[m] * finv[n] % mod * finv[m - n] % mod;}ll A(int n, int m){if(m n || n 0 || m 0) return 0;return 1ll * fact[m] * finv[m - n] % mod;}ll llC(int n, int m){//求解不进行取余mod的可能爆longlong的C(n,m)值;if(m n || n 0 || m 0) return 0;swap(n, m);long long sum 1;if (m n - m) m n - m;n n - m 1;for (int i 1; i m; i ){sum * n ;sum / i;}return sum;}};templatesize_t sizestruct Prime{int con[size], tot 0, vis[size];void init(){for(int i 2; i (int)size; i ){if(vis[i] 0) con[ tot] i;for(int j 1; j tot i * con[j] (int)size; j ){vis[i * con[j]] con[j];if(i % con[j] 0) break;}}}};// Road maxn e; // 无向图 * 2// Combinatorial_number N comb;// Prime N prime;//题意给定一个字符串由()?组成可以将?替换成(或)求能组成满足()的个数int n,m;int a[N];int dp[3010][3010]; //dp[i][j]表示在i位置,(记作1)表示-1j表示前缀和void solve(){string s;cins;ns.size();s s;dp[0][0]1;int Max0;fr(i,1,n){fr(j,0,i){if(s[i](){if(j)dp[i][j]dp[i-1][j-1];}else if(s[i])){dp[i][j]dp[i-1][j1];}else {if(!j) dp[i][j]dp[i-1][j1];else dp[i][j](dp[i-1][j1]dp[i-1][j-1])%mod;}}}coutdp[n][0]\n;}signed main(){int t1;// cint;while(t--) solve();return 0;}
