結果
問題 | No.1243 約数加算 |
ユーザー |
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提出日時 | 2020-10-02 22:56:27 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 7 ms / 2,000 ms |
コード長 | 6,828 bytes |
コンパイル時間 | 1,088 ms |
コンパイル使用メモリ | 89,804 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-07-17 22:59:04 |
合計ジャッジ時間 | 1,780 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 9 |
コンパイルメッセージ
main.cpp: In function 'long long int keta(long long int)': main.cpp:36:1: warning: control reaches end of non-void function [-Wreturn-type] 36 | } | ^ main.cpp: In function 'long long int gcd(long long int, long long int)': main.cpp:50:1: warning: control reaches end of non-void function [-Wreturn-type] 50 | } | ^ main.cpp: In function 'long long int lcm(long long int, long long int)': main.cpp:63:1: warning: control reaches end of non-void function [-Wreturn-type] 63 | } | ^
ソースコード
#include<iostream>#include<algorithm>#include<cmath>#include<map>#include<stdio.h>#include<vector>#include<queue>#include<math.h>#include<deque>using namespace std;#define int long long#define rep(s,i,n) for(int i=s;i<n;i++)#define c(n) cout<<n<<endl;#define ic(n) int n;cin>>n;#define sc(s) string s;cin>>s;#define mod 1000000007#define inf 1000000000000000007#define f first#define s second#define mini(c,a,b) *min_element(c+a,c+b)#define maxi(c,a,b) *max_element(c+a,c+b)#define pi 3.141592653589793238462643383279#define e_ 2.718281828459045235360287471352#define P pair<int,int>#define upp(a,n,x) upper_bound(a,a+n,x)-a;#define low(a,n,x) lower_bound(a,a+n,x)-a;#define UF UnionFind//printf("%.12Lf\n",);int keta(int x) {rep(0, i, 30) {if (x < 10) {return i + 1;}x = x / 10;}}int gcd(int x, int y) {if (x == 0 || y == 0)return x + y;int aa = x, bb = y;rep(0, i, 1000) {aa = aa % bb;if (aa == 0) {return bb;}bb = bb % aa;if (bb == 0) {return aa;}}}int lcm(int x, int y) {int aa = x, bb = y;rep(0, i, 1000) {aa = aa % bb;if (aa == 0) {return x / bb * y;}bb = bb % aa;if (bb == 0) {return x / aa * y;}}}bool p(int x) {if (x == 1)return false;rep(2, i, sqrt(x) + 1) {if (x % i == 0 && x != i) {return false;}}return true;}int max(int a, int b) {if (a >= b)return a;else return b;}string maxst(string s, string t) {int n = s.size();int m = t.size();if (n > m)return s;else if (n < m)return t;else {rep(0, i, n) {if (s[i] > t[i])return s;if (s[i] < t[i])return t;}return s;}}int min(int a, int b) {if (a >= b)return b;else return a;}int n2[61];int nis[61];int nia[61];int mody[61];int nn;int com(int n, int y) {int ni = 1;for (int i = 0;i < 41;i++) {n2[i] = ni;ni *= 2;}int bunsi = 1, bunbo = 1;rep(0, i, y)bunsi = (bunsi * (n - i)) % mod;rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod;mody[0] = bunbo;rep(1, i, 41) {bunbo = (bunbo * bunbo) % mod;mody[i] = bunbo;}rep(0, i, 41)nis[i] = 0;nn = mod - 2;for (int i = 40;i >= 0;i -= 1) {if (nn > n2[i]) {nis[i]++;nn -= n2[i];}}nis[0]++;rep(0, i, 41) {if (nis[i] == 1) {bunsi = (bunsi * mody[i]) % mod;}}return bunsi;}int gyakugen(int n, int y) {int ni = 1;for (int i = 0;i < 41;i++) {n2[i] = ni;ni *= 2;}mody[0] = y;rep(1, i, 41) {y = (y * y) % mod;mody[i] = y;}rep(0, i, 41)nis[i] = 0;nn = mod - 2;for (int i = 40;i >= 0;i -= 1) {if (nn > n2[i]) {nis[i]++;nn -= n2[i];}}nis[0]++;rep(0, i, 41) {if (nis[i] == 1) {n = (n * mody[i]) % mod;}}return n;}int yakuwa(int n) {int sum = 0;rep(1, i, sqrt(n + 1)) {if (n % i == 0)sum += i + n / i;if (i * i == n)sum -= i;}return sum;}int poow(int y, int n) {if (n == 0)return 1;n -= 1;int ni = 1;for (int i = 0;i < 61;i++) {n2[i] = ni;ni *= 2;}int yy = y;mody[0] = yy;rep(1, i, 61) {yy = (yy * yy) % mod;mody[i] = yy;}rep(0, i, 61)nis[i] = 0;nn = n;for (int i = 60;i >= 0;i -= 1) {if (nn >= n2[i]) {nis[i]++;nn -= n2[i];}}rep(0, i, 61) {if (nis[i] == 1) {y = (y * mody[i]) % mod;}}return y;}int minpow(int x, int y) {int sum = 1;rep(0, i, y)sum *= x;return sum;}int ketawa(int x, int sinsuu) {int sum = 0;rep(0, i, 100)sum += (x % poow(sinsuu, i + 1)) / (poow(sinsuu, i));return sum;}int sankaku(int a) {return a * (a + 1) / 2;}int sames(int a[1111111], int n) {int ans = 0;rep(0, i, n) {if (a[i] == a[i + 1]) {int j = i;while (a[j + 1] == a[i] && j <= n - 2)j++;ans += sankaku(j - i);i = j;}}return ans;}using Graph = vector<vector<int>>;int oya[114514];int depth[114514];void dfs(const Graph& G, int v, int p, int d) {depth[v] = d;oya[v] = p;for (auto nv : G[v]) {if (nv == p) continue; // nv が親 p だったらダメdfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ}}/*int H=10,W=10;char field[10][10];char memo[10][10];void dfs(int h, int w) {memo[h][w] = 'x';// 八方向を探索for (int dh = -1; dh <= 1; ++dh) {for (int dw = -1; dw <= 1; ++dw) {if(abs(0-dh)+abs(0-dw)==2)continue;int nh = h + dh, nw = w + dw;// 場外アウトしたり、0 だったりはスルーif (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;if (memo[nh][nw] == 'x') continue;// 再帰的に探索dfs(nh, nw);}}}*/int XOR(int a, int b) {if (a == 0 || b == 0) {return a + b;}int ni = 1;rep(0, i, 41) {n2[i] = ni;ni *= 2;}rep(0, i, 41)nis[i] = 0;for (int i = 40;i >= 0;i -= 1) {if (a >= n2[i]) {nis[i]++;a -= n2[i];}if (b >= n2[i]) {nis[i]++;b -= n2[i];}}int sum = 0;rep(0, i, 41)sum += (nis[i] % 2 * n2[i]);return sum;}int string_int(string s){int sum=0;int n=s.size();int t=1;rep(0,i,n){sum+=t*(s[n-i-1]-'0');t*=10;}return sum;}//int ma[1024577][21];//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;struct UnionFind {vector<int> par; // par[i]:iの親の番号 (例) par[3] = 2 : 3の親が2UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化for (int i = 0; i < N; i++) par[i] = i;}int root(int x) { // データxが属する木の根を再帰で得る:root(x) = {xの木の根}if (par[x] == x) return x;return par[x] = root(par[x]);}void unite(int x, int y) { // xとyの木を併合int rx = root(x); //xの根をrxint ry = root(y); //yの根をryif (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのままpar[rx] = ry; //xとyの根が同じでない(=同じ木にない)時:xの根rxをyの根ryにつける}bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返すint rx = root(x);int ry = root(y);return rx == ry;}};int two[70];signed main(){ic(q)two[0]=1;rep(1,i,62)two[i]=two[i-1]*2;while(q>0){q--;ic(a) ic(b)if(b-a==1){c(1)c(1)continue;}vector<int> ans;int A[70],B[70];int k=1;rep(0,i,61){A[i]=a%(k*2)/k;B[i]=b%(k*2)/k;k*=2;}int c[70];bool t=true;for(int i=60;i>=0;i-=1){if(B[i]>A[i]&&t){t=false;c[i]=1;}else{if(t)c[i]=B[i];else c[i]=0;}}int p=0;k=1;rep(0,i,61){p+=c[i]*k;k*=2;}int s=a;rep(0,i,61){bool m=true;rep(1,j,61){if(s%two[j]>0&&s<p&&m){ans.push_back(two[j-1]);s+=two[j-1];m=false;}}}for(int i=61;i>=0;i-=1){if(B[i]==1&&c[i]==0){ans.push_back(two[i]);s+=two[i];}}c(ans.size());rep(0,i,ans.size())cout<<ans[i]<<" ";cout<<endl;}}