hdu 5288 OO’s Sequence
题意：f (L,R)表示下标i在区间内且除了a[i]本身，区间内的其他数都不能整除a[i]，这样的数的个数。现在给出n，和a[1]~a[n]，求∑i=1~n∑j=i~n f (i,j) mod （10^9+7）。
做法：定义L[i]表示在a[i]左边第一个整除的数的下标，R[i]表示在a[i]右边第一个整除的数的下标，则ans += (R[i] - i) * (i - L[i])。pre[i]表示 数i上一次出现的位置
求R[i]：从左到右访问每一个位置i，并用pre[i]标记，对于每一个数a[i]，看其倍数是否在其之前出现，如果已经出现，那么已经出现的那个数，它的R[i]就是min(R[i],i)。同理求出L[i]。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <cmath> 6 #include <algorithm> 7 #include <queue> 8 #include <vector> 9 10 using namespace std; 11 12 #define LL long long 13 #define eps 1e-8 14 #define inf 0x3f3f3f3f 15 #define N 200010 16 #define Max 10010 17 #define M 400020 18 #define mod 1000000007 19 20 int L[N], R[N],a[N], pre[N]; 21 int main(){ 22 int n; 23 while(scanf("%d", &n) != EOF){ 24 memset(pre, 0, sizeof pre); 25 for(int i = 1; i <= n; ++i){ 26 scanf("%d", &a[i]); 27 L[i] = 0, R[i] = n + 1; 28 for(int j = a[i]; j < Max; j += a[i]){ 29 if(pre[j] && R[pre[j]] == n + 1) R[pre[j]] = i; 30 } 31 pre[a[i]] = i; 32 } 33 memset(pre, 0, sizeof pre); 34 for(int i = n; i > 0; --i){ 35 for(int j = a[i]; j < Max; j += a[i]){ 36 if(pre[j] && L[pre[j]] == 0) L[pre[j]] = i; 37 } 38 pre[a[i]] = i; 39 } 40 LL ans = 0; 41 for(int i = 1; i <= n; ++i){ 42 ans += (LL)(i - L[i]) * (LL)(R[i] - i); 43 ans %= mod; 44 } 45 printf("%I64d ", ans); 46 } 47 return 0; 48 }
hdu 5289 Assignment
题意：问有多少个区间满足 区间最大 - 区间最小 < k。
做法：枚举左端点，二分右端点，用st算法求区间最大最小。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <cmath> 6 #include <algorithm> 7 #include <queue> 8 #include <vector> 9 10 using namespace std; 11 12 #define LL long long 13 #define eps 1e-8 14 #define inf 0x3f3f3f3f 15 #define N 100010 16 #define M 400020 17 #define mod 1000000007 18 #define lson l, m, rt << 1 19 #define rson m + 1, r, rt << 1 | 1 20 21 int mi[N][20], mx[N][20], a[N], n; 22 void RMQ_init(){ 23 for(int i = 1; i <= n; ++i) mi[i][0] = mx[i][0] = a[i]; 24 for(int j= 1; (1 << j) <= n + 1; ++j) 25 for(int i = 1; i + (1 << j) - 1 <= n; ++i) 26 mi[i][j] = min(mi[i][j-1], mi[i + (1 << (j-1))][j-1]), 27 mx[i][j] = max(mx[i][j-1], mx[i + (1 << (j-1))][j-1]); 28 } 29 void query(int L, int R, int &Max, int &Min){ 30 int k = 0; 31 while((1 << (k+1)) <= R - L + 1) k++; 32 Max = max(mx[L][k], mx[R - (1<<k) + 1][k]); 33 Min = min(mi[L][k], mi[R - (1<<k) + 1][k]); 34 } 35 int read_int(){ 36 int x = 0; char ch = getchar(); 37 while(ch < '0' || ch > '9') ch = getchar(); 38 while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); 39 return x; 40 } 41 int main(){ 42 int cas, k; 43 scanf("%d", &cas); 44 while(cas--){ 45 scanf("%d%d", &n, &k); 46 for(int i = 1; i <= n; ++i) 47 a[i] = read_int(); 48 RMQ_init(); 49 LL ans = 0; 50 for(int i = 1; i <= n; ++i){ 51 int Max, Min, L = i, R = n; 52 while(L < R){ 53 int mid = (L + R) >> 1; 54 Min = inf, Max = -inf; 55 query(i, mid, Max, Min); 56 if(Max - Min >= k) 57 R = mid; 58 else L = mid + 1; 59 } 60 Min = inf, Max = -inf; 61 query(i, L, Max, Min); 62 if(Max - Min >= k) 63 L--; 64 ans += (L - i + 1); 65 } 66 printf("%I64d ", ans); 67 } 68 return 0; 69 }
hdu 5191 Candy Distribution
题意：有n种糖果，每种有ai个，要分给两个人，每个糖果可以分或者不分，问你有多少种方案，使得两个人分得的糖果一样多。
做法：dp[i][j]表示前 i 种糖果，第一个人比第二个人多 j 个糖果的方案数。考虑dp[i][j] 对 dp[i+1][..]的影响，则设分给第一个人 x 个，第二个人 y 个，只要 x + y <= a[i+1]，都在计算范围内，则dp[i][j] 对 dp[i+1][j]的贡献有 a[i+1] / 2次，对 dp[i+1][j+1]的贡献是(a[i+1] - 1)/2次。dp[i][j]对dp[i+1][..]的贡献按照a[i+1]的奇偶性不同会呈现一种类似等差数列的方式。然后算一算就好了。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<string> 5 #include<cstring> 6 #include<algorithm> 7 #include<set> 8 using namespace std; 9 10 11 #define inf 0x3f3f3f3f 12 #define eps 1e-6 13 #define MP make_pair 14 #define LL long long 15 #define N 400010 16 #define M 200020 17 #define mod 1000000007 18 19 int a[220]; 20 int dp[2][N<<1], p[N<<1]; 21 void add(int &x, int y){ 22 x = x + y; 23 if(x >= mod) x -= mod; 24 if(x < 0) x += mod; 25 } 26 void cal(int i, int v, int x){ 27 if(v & 1){ 28 add(p[i-v], x), add(p[i+1], -x); 29 add(p[i+3], -x), add(p[i+v+4], x); 30 31 add(p[i-v+1], x), add(p[i+2], mod-2*x); 32 add(p[i+v+3], x); 33 } 34 else{ 35 add(p[i-v], x), add(p[i+2], mod-2*x); 36 add(p[i+v+4], x); 37 38 add(p[i-v+1], x), add(p[i+1], -x); 39 add(p[i+3], -x), add(p[i+v+3], x); 40 } 41 } 42 int main(){ 43 int cas; 44 scanf("%d", &cas); 45 while(cas--){ 46 int n; 47 scanf("%d", &n); 48 for(int i = 0; i < n; ++i) 49 scanf("%d", &a[i]); 50 memset(dp, 0, sizeof dp); 51 memset(p, 0, sizeof p); 52 int all = 0, cur = 1, pre = 0; 53 dp[0][N] = 1; 54 for(int i = 0; i < n; ++i){ 55 all += a[i]; 56 for(int j = N - all - 2; j <= N + all + 2; ++j) 57 dp[cur][j] = p[j] = 0; 58 for(int j = N - all; j <= N + all; ++j){ 59 if(dp[pre][j] == 0) continue; 60 cal(j, a[i], dp[pre][j]); 61 } 62 int val = 0, sum = 0; 63 for(int j = N - all; j <= N + all; j += 2){ 64 add(val, p[j]); 65 add(sum, val); 66 dp[cur][j] = sum; 67 } 68 val = sum = 0; 69 for(int j = N - all + 1; j < N + all + 1 ; j += 2){ 70 add(val, p[j]); 71 add(sum, val); 72 dp[cur][j] = sum; 73 } 74 cur ^= 1, pre ^= 1; 75 } 76 printf("%d ", dp[pre][N]); 77 } 78 return 0; 79 }
hdu 5294 Tricks Device
做法：求最短路，总边数 - 最少边数的最短路 的边数 即为第二个问答案。在所有的最短路的边上 构每条边流量为1的新图，求最大流即为第一个问的答案
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <cmath> 6 #include <algorithm> 7 #include <queue> 8 #include <vector> 9 10 using namespace std; 11 12 #define LL long long 13 #define eps 1e-8 14 #define inf 0x3f3f3f3f 15 #define N 5010 16 #define M 120020 17 #define mod 1000000007 18 19 struct edge { 20 int u, v, cap, flow, nxt; 21 void set(int _u, int _v, int _cap, int _flow, int _nxt) { 22 u = _u, v = _v, cap = _cap, flow = _flow, nxt = _nxt; 23 } 24 }; 25 26 struct dinic { 27 int s, t, fst[N], cc, d[N], cur[N]; 28 edge e[M]; 29 void init() { 30 memset(fst, -1, sizeof(fst)); 31 cc = 0; 32 } 33 void add(int u, int v, int cap) { 34 e[cc].set(u, v, cap, 0, fst[u]), fst[u] = cc++; 35 e[cc].set(v, u, 0, 0, fst[v]), fst[v] = cc++; 36 } 37 bool bfs() { 38 memset(d, -1, sizeof(d)); 39 queue<int> q; 40 d[s] = 0, q.push(s); 41 while(!q.empty()) { 42 int u = q.front(); q.pop(); 43 for(int i = fst[u]; ~i; i = e[i].nxt) { 44 int v = e[i].v; 45 if(d[v] == -1 && e[i].cap > e[i].flow) { 46 d[v] = d[u] + 1; 47 q.push(v); 48 } 49 } 50 } 51 return d[t] != -1; 52 } 53 int dfs(int x, int a) { 54 if(x == t || a == 0) return a; 55 int f, flow = 0; 56 for(int &i = cur[x]; ~i; i = e[i].nxt) { 57 if(i == inf) i = fst[x]; 58 if(i == -1) break; 59 int v = e[i].v; 60 if(d[v] == d[x] + 1 && (f = dfs(v, min(a, e[i].cap - e[i].flow))) > 0) { 61 a -= f, e[i ^ 1].flow -= f; 62 e[i].flow += f, flow += f; 63 if(!a) break; 64 } 65 } 66 return flow; 67 } 68 int go(int s, int t) { 69 this -> s = s, this -> t = t; 70 int flow = 0; 71 while(bfs()) { 72 memset(cur, 0x3f, sizeof(cur)); 73 flow += dfs(s, inf); 74 } 75 return flow; 76 } 77 }go; 78 int fst[N], vv[M], nxt[M], cost[M], e; 79 void init(){ 80 memset(fst, -1, sizeof fst); 81 e = 0; 82 } 83 void add(int u, int v, int c){ 84 vv[e] = v, nxt[e] = fst[u], cost[e] = c, fst[u] = e++; 85 } 86 int d[N], n; 87 bool vis[N]; 88 void spfa(){ 89 memset(d, 0x3f, sizeof d); 90 memset(vis, 0, sizeof vis); 91 queue<int> q; 92 q.push(1); d[1] = 0, vis[1] = 1; 93 while(!q.empty()){ 94 int u = q.front(); q.pop(); 95 vis[u] = 0; 96 for(int i = fst[u]; ~i; i = nxt[i]){ 97 int v = vv[i], c = cost[i]; 98 if(d[v] > d[u] + c){ 99 d[v] = d[u] + c; 100 if(!vis[v]) vis[v] = 1, q.push(v); 101 } 102 } 103 } 104 } 105 vector<int> g[N]; 106 int bfs(){ 107 memset(d, 0x3f, sizeof d); 108 queue<int> q; 109 d[1] = 0, vis[1] = 1; 110 q.push(1); 111 while(!q.empty()){ 112 int u = q.front(); q.pop(); 113 for(int i = 0; i < g[u].size(); ++i){ 114 int v = g[u][i]; 115 if(!vis[v]) 116 d[v] = d[u] + 1, vis[v] = 1, q.push(v); 117 } 118 } 119 return d[n]; 120 } 121 int main(){ 122 int m; 123 while(scanf("%d%d", &n, &m) != EOF){ 124 init(); 125 for(int i = 0; i < m; ++i){ 126 int u, v, c; 127 scanf("%d%d%d", &u, &v, &c); 128 add(u, v, c), add(v, u, c); 129 } 130 spfa(); 131 go.init(); 132 for(int i = 1; i <= n; ++i) 133 g[i].clear(); 134 for(int i = 1; i <= n; ++i) 135 for(int j = fst[i]; ~j; j = nxt[j]){ 136 int v = vv[j], c = cost[j]; 137 if(d[v] == d[i] + c){ 138 go.add(i, v, 1); 139 g[i].push_back(v); 140 } 141 } 142 printf("%d %d ", go.go(1, n), m - bfs()); 143 } 144 return 0; 145 }
hdu 5295 Unstable
题意：给出AB、BC、CD、DA、EF的长度，求一个合法的A、B、C、D的坐标。
一张不知道哪里来的图。
做法：固定BC，然后求出A'和G点的坐标，再求出A和D的坐标。。注意有精度误差。eps最好是1e-5或者1e-4。还有注意圆重合、内切、外切的情况。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <string> 5 #include <cstring> 6 #include <algorithm> 7 #include <set> 8 using namespace std; 9 10 11 #define inf 0x3f3f3f3f 12 #define eps 1e-4 13 #define pii pair<int,int> 14 #define MP make_pair 15 #define LL long long 16 #define N 100010 17 #define M 200020 18 #define mod 1000000007 19 20 int dcmp( double x ){ 21 if( fabs( x ) < eps ) return 0; 22 return x < 0 ? -1 : 1; 23 } 24 struct point{ 25 double x, y; 26 point( double x = 0, double y = 0 ) : x(x), y(y) {} 27 point operator + ( const point &b ) const{ 28 return point( x + b.x, y + b.y ); 29 } 30 point operator - ( const point &b ) const{ 31 return point( x - b.x, y - b.y ); 32 } 33 point operator * ( const double &k ) const{ 34 return point( x * k, y * k ); 35 } 36 point operator / ( const double &k ) const{ 37 return point( x / k, y / k ); 38 } 39 bool operator < ( const point &b ) const{ 40 return dcmp( x - b.x ) < 0 || dcmp( x - b.x ) == 0 && dcmp( y - b.y ) < 0; 41 } 42 bool operator == ( const point &b ) const{ 43 return dcmp( x - b.x ) == 0 && dcmp( y - b.y ) == 0; 44 } 45 double len(){ 46 return sqrt( x * x + y * y ); 47 } 48 void out(){ 49 printf("%.8lf %.8lf ", x, y); 50 } 51 }; 52 void circle_cross( point a, double ra, point b, double rb, point &v1, point &v2 ){ 53 if(a == b && dcmp(ra - rb) == 0){ 54 v1 = a + point(0, ra); 55 v2 = a - point(0, ra); 56 return ; 57 } 58 double d = ( a - b ).len(); 59 if(dcmp(d + min(ra, rb) - max(ra, rb)) == 0){ 60 if(ra > rb) swap(ra, rb), swap(a, b); 61 v1 = v2 = a + (a - b) / d * ra; 62 return ; 63 } 64 if(dcmp(d - ra - rb) == 0){ 65 v1 = v2 = a + (b - a) / d * ra; 66 return ; 67 } 68 double da = ( ra * ra + d * d - rb * rb ) / ( 2 * ra * d ); 69 double aa = atan2( (b-a).y, (b-a).x ); 70 double rad = acos( da ); 71 v1 = point( a.x + cos( aa - rad ) * ra, a.y + sin( aa - rad ) * ra ); 72 v2 = point( a.x + cos( aa + rad ) * ra, a.y + sin( aa + rad ) * ra ); 73 } 74 75 double ab, bc, cd, da, ef; 76 point a, b, c, d; 77 void solve(){ 78 c = point(0, 0), b = point(bc, 0); 79 point a2, tmp; 80 circle_cross(c, da, b, 2*ef, a2, tmp); 81 point g = a2 - c + b; 82 circle_cross(c, cd, g, ab, d, tmp); 83 a = d - g + b; 84 a.out(), b.out(), c.out(), d.out(); 85 } 86 int main(){ 87 //freopen("tt.txt", "r", stdin); 88 int cas, kk = 1; 89 scanf("%d", &cas); 90 while(cas--){ 91 printf("Case #%d: ", kk++); 92 scanf("%lf%lf%lf%lf%lf", &ab, &bc, &cd, &da, &ef); 93 solve(); 94 } 95 return 0; 96 }
hdu5297 Y sequence
题意：给你n 和 r。求出从1开始，去除掉所有能够写成a^b（2 <= b <= r）的数后的第n个数的值。
做法：先预处理出5 - 60以内的所有素数次方。对于每次询问，二分答案。对于2次方和3次方，可以用容斥去掉。然后再处理其他<=r 的次方 的数。有几个坑点。
1.2^2^5、2^3^5、3^2^5、3^3^5……2^2^7、2^3^7、3^2^7、3^3^7……等在预处理的时候不要算进去，因为会在用容斥处理了2次方和3次方的时候去掉了。
2.2^5^7、2^5^11、3^5^7有可能被减去两次，要加回来。
3.二分完答案后一定要再check一下，有可能这个答案刚好是一个要被删的数。
我是按照题解说的做的。还有用莫比乌斯做的（看不懂），以及直接容斥过的（不会精度误差吗）。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <cstring> 6 #include <string> 7 #include <set> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-8 13 #define N 210 14 #define M 20020 15 16 int prime[N], tot; 17 bool vis[N]; 18 const LL Max = 2 * 1e18, Lar = 1e18; 19 const LL a35 = 1LL << 35, a55 = 1LL << 55, b35 = 50031545098999707LL; 20 LL a[30][5000]; 21 LL qpow(LL x, int k){ 22 LL ret = 1; 23 while(k){ 24 if(k & 1){ 25 double val = (double)ret * x; 26 if(val > Max) 27 return Max + 1; 28 ret *= x; 29 } 30 x *= x; 31 k >>= 1; 32 } 33 return ret; 34 } 35 bool check2(int j){ 36 int tmp = sqrt(j+0.5); 37 if(tmp * tmp == j) return 1; 38 tmp = pow(j +0.5, 1.0/3); 39 for(int i = tmp - 10; i < tmp + 11; ++i) 40 if(i * i * i == j) return 1; 41 return 0; 42 } 43 int b[20]; 44 void init(){ 45 for(int i = 2; i < 60; ++i){ 46 if(!vis[i]) prime[++tot] = i; 47 for(int j = i + i; j < 60; j +=i) 48 vis[j] = 1; 49 } 50 for(int i = 3; i <= tot; ++i){ 51 int cnt = 0, j = 2; 52 while(1){ 53 LL tmp = qpow(j, prime[i]); 54 if(tmp > Max) break; 55 j++, a[i][cnt++] = tmp; 56 while(check2(j)) j++; 57 } 58 b[i] = cnt; 59 } 60 } 61 62 LL n; int r; 63 bool check(LL x){ 64 if(x < 60 && !vis[x]) return 0; 65 LL ret = x, cnt = 0; 66 while(ret % 2 == 0) ret /= 2, cnt++; 67 if(ret == 1 && cnt <= r) return 1; 68 ret = x, cnt = 0; 69 while(ret % 3 == 0) ret /= 3, cnt++; 70 if(ret == 1 && cnt <= r) return 1; 71 for(int i = 3; i <= tot; ++i){ 72 if(prime[i] > r) break; 73 int L = 0, R = b[i]; 74 while(L < R){ 75 int mid = (L + R) >> 1; 76 if(a[i][mid] < n) 77 L = mid + 1; 78 else R = mid; 79 } 80 if(a[i][L] == x) return 1; 81 } 82 return 0; 83 } 84 LL x, y, z; 85 LL f(LL val){ 86 LL temp = val; 87 x = pow(val+0.5, 1.0/2), y = pow(val+0.5, 1.0/3), z = pow(val+0.5, 1.0/6); 88 if(2 <= r) temp -= x; 89 if(3 <= r) temp -= y, temp += z; 90 for(int i = 3; i <= tot; ++i){ 91 if(prime[i] > r) break; 92 int L = upper_bound(a[i], a[i]+b[i], val) - a[i]; 93 if(L == 0) break; 94 temp -= L; 95 } 96 if(val >= a35 && r >= 7) temp++; 97 if(val >= a55 && r >= 11) temp++; 98 if(val >= b35 && r >= 7) temp++; 99 return temp; 100 } 101 int main(){ 102 //freopen("1010.in", "r", stdin); 103 //freopen("tt.txt", "w", stdout); 104 init(); 105 int cas; 106 scanf("%d", &cas); 107 while(cas--){ 108 109 scanf("%I64d%d", &n, &r); 110 LL L = n, R = n + 10000000000; 111 while(L < R){ 112 LL mid = (L + R) >> 1; 113 if(f(mid) < n) 114 L = mid + 1; 115 else R = mid; 116 } 117 //cout <<L << endl; 118 while(check(L)) 119 L++; 120 printf("%I64d ", L); 121 } 122 return 0; 123 }
hdu5298 Solid Geometry Homework
题意：给一些平面和球的方程，要进行染色（红色和黄色）、以及一些要染成黄色的点（点所在的区域要染成黄色），有q个询问，每次询问一个点，问你这个点在的区域是什么颜色。
做法：首先，如果没有给出要染成黄色的点，那么询问的答案就是both。如果给出了黄色的点，那么染色方案就确定了，接下来输入要染成黄色的点 有没有冲突，有冲突就impossible。没有就可以回答询问了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <cstring> 6 #include <string> 7 #include <set> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-8 13 #define N 210 14 #define M 20020 15 16 int prime[N], tot; 17 bool vis[N]; 18 const LL Max = 2 * 1e18, Lar = 1e18; 19 const LL a35 = 1LL << 35, a55 = 1LL << 55, b35 = 50031545098999707LL; 20 LL a[30][5000]; 21 LL qpow(LL x, int k){ 22 LL ret = 1; 23 while(k){ 24 if(k & 1){ 25 double val = (double)ret * x; 26 if(val > Max) 27 return Max + 1; 28 ret *= x; 29 } 30 x *= x; 31 k >>= 1; 32 } 33 return ret; 34 } 35 bool check2(int j){ 36 int tmp = sqrt(j+0.5); 37 if(tmp * tmp == j) return 1; 38 tmp = pow(j +0.5, 1.0/3); 39 for(int i = tmp - 10; i < tmp + 11; ++i) 40 if(i * i * i == j) return 1; 41 return 0; 42 } 43 int b[20]; 44 void init(){ 45 for(int i = 2; i < 60; ++i){ 46 if(!vis[i]) prime[++tot] = i; 47 for(int j = i + i; j < 60; j +=i) 48 vis[j] = 1; 49 } 50 for(int i = 3; i <= tot; ++i){ 51 int cnt = 0, j = 2; 52 while(1){ 53 LL tmp = qpow(j, prime[i]); 54 if(tmp > Max) break; 55 j++, a[i][cnt++] = tmp; 56 while(check2(j)) j++; 57 } 58 b[i] = cnt; 59 } 60 } 61 62 LL n; int r; 63 bool check(LL x){ 64 if(x < 60 && !vis[x]) return 0; 65 LL ret = x, cnt = 0; 66 while(ret % 2 == 0) ret /= 2, cnt++; 67 if(ret == 1 && cnt <= r) return 1; 68 ret = x, cnt = 0; 69 while(ret % 3 == 0) ret /= 3, cnt++; 70 if(ret == 1 && cnt <= r) return 1; 71 for(int i = 3; i <= tot; ++i){ 72 if(prime[i] > r) break; 73 int L = 0, R = b[i]; 74 while(L < R){ 75 int mid = (L + R) >> 1; 76 if(a[i][mid] < n) 77 L = mid + 1; 78 else R = mid; 79 } 80 if(a[i][L] == x) return 1; 81 } 82 return 0; 83 } 84 LL x, y, z; 85 LL f(LL val){ 86 LL temp = val; 87 x = pow(val+0.5, 1.0/2), y = pow(val+0.5, 1.0/3), z = pow(val+0.5, 1.0/6); 88 if(2 <= r) temp -= x; 89 if(3 <= r) temp -= y, temp += z; 90 for(int i = 3; i <= tot; ++i){ 91 if(prime[i] > r) break; 92 int L = upper_bound(a[i], a[i]+b[i], val) - a[i]; 93 if(L == 0) break; 94 temp -= L; 95 } 96 if(val >= a35 && r >= 7) temp++; 97 if(val >= a55 && r >= 11) temp++; 98 if(val >= b35 && r >= 7) temp++; 99 return temp; 100 } 101 int main(){ 102 //freopen("1010.in", "r", stdin); 103 //freopen("tt.txt", "w", stdout); 104 init(); 105 int cas; 106 scanf("%d", &cas); 107 while(cas--){ 108 109 scanf("%I64d%d", &n, &r); 110 LL L = n, R = n + 10000000000; 111 while(L < R){ 112 LL mid = (L + R) >> 1; 113 if(f(mid) < n) 114 L = mid + 1; 115 else R = mid; 116 } 117 //cout <<L << endl; 118 while(check(L)) 119 L++; 120 printf("%I64d ", L); 121 } 122 return 0; 123 }