标签:iss number first query 欧拉 win main miss printf
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 738 Accepted Submission(s): 333
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<math.h> 6 #include<queue> 7 #include<vector> 8 #include<stack> 9 #include<set> 10 using namespace std; 11 typedef long long LL; 12 int ans[100000]; 13 int mul[100000]; 14 typedef struct node 15 { 16 int l; 17 int r; 18 int id; 19 } ss; 20 ss ask[100000]; 21 bool cmp1(node p,node q) 22 { 23 return p.l < q.l; 24 } 25 bool cmp2(node p,node q) 26 { 27 return p.r < q.r; 28 } 29 bool prime[30000]; 30 int prime_table[30000]; 31 vector<int>vec[30000]; 32 int cnt[20005]; 33 LL answ[30000]; 34 int oula[20005]; 35 void _slove_mo(int n,int m); 36 int main(void) 37 { 38 int n,m; 39 int T; 40 int __ca = 0; 41 int cn = 0; 42 mul[1] = 1; 43 int i,j; 44 memset(prime,0,sizeof(prime)); 45 for(i = 0; i <= 20000; i++) 46 oula[i] = i; 47 for(i = 2; i <= 20000; i++) 48 { 49 if(!prime[i]) 50 { 51 prime_table[cn++] = i; 52 mul[i] = -1; 53 } 54 for(j = 0; j < cn&&(i*prime_table[j]<=20000); j++) 55 { 56 if(i%prime_table[j]) 57 { 58 prime[i*prime_table[j]] = true; 59 mul[i*prime_table[j]] = -mul[i]; 60 } 61 else 62 { 63 prime[i*prime_table[j]] = true; 64 mul[i*prime_table[j]] = 0; 65 break; 66 } 67 } 68 }//printf("%d\n",cn); 69 for(i = 0; i < cn; i++) 70 { 71 for(j = 1; j*prime_table[i]<=20000; j++) 72 { 73 oula[j*prime_table[i]]/=prime_table[i]; 74 oula[j*prime_table[i]]*=(prime_table[i]-1); 75 } 76 } 77 for(i = 1; i <= 20000; i++) 78 { 79 for(j = 1; j <= sqrt(i); j++) 80 { 81 if(i%j==0) 82 { 83 vec[i].push_back(j); 84 if(i/j != j) 85 vec[i].push_back(i/j); 86 } 87 } 88 }scanf("%d",&T); 89 while(T--) 90 { 91 ++__ca; memset(cnt,0,sizeof(cnt)); 92 scanf("%d",&n); 93 for(i = 1; i <= n; i++) 94 { 95 scanf("%d",&ans[i]); 96 } 97 scanf("%d",&m); 98 for(i = 0; i < m; i++) 99 { 100 scanf("%d %d",&ask[i].l,&ask[i].r); 101 ask[i].id = i; 102 } 103 sort(ask,ask+m,cmp1); 104 int id = 0; 105 int ak = sqrt(1.0*n)+1; 106 int v = ak; 107 for(i = 0; i < m; i++) 108 { 109 if(ask[i].l > v) 110 { 111 v += ak; 112 sort(ask+id,ask+i,cmp2); 113 id = i; 114 } 115 } 116 sort(ask+id,ask+m,cmp2); 117 _slove_mo(n,m); 118 printf("Case #%d:\n",__ca); 119 for(i = 0; i < m; i++) 120 printf("%lld\n",answ[i]); 121 122 }return 0; 123 } 124 void _slove_mo(int n,int m) 125 { 126 int i,j; 127 LL sum = 0; 128 int xl = ask[0].l; 129 int xr = ask[0].r; 130 for(i = xl; i <= xr; i++) 131 { 132 for(j = 0; j < vec[ans[i]].size(); j++) 133 { int x = vec[ans[i]][j]; 134 sum = sum + (LL)oula[x]*(LL)cnt[x]; 135 cnt[x]++; 136 } 137 } 138 answ[ask[0].id] = sum; 139 for(i = 1; i < m; i++) 140 { 141 while(xl < ask[i].l) 142 { 143 int y = ans[xl]; 144 for(j = 0; j < vec[y].size(); j++) 145 { 146 int x = vec[y][j]; 147 sum -= (LL)oula[x]*(LL)(--cnt[x]); 148 } 149 xl++; 150 } 151 while(xl > ask[i].l) 152 { 153 xl--; 154 int y = ans[xl]; 155 for(j = 0; j < vec[y].size(); j++) 156 { 157 int x = vec[y][j]; 158 sum += (LL)oula[x]*(LL)(cnt[x]++); 159 } 160 } 161 while(xr > ask[i].r) 162 { 163 int y = ans[xr]; 164 for(j = 0; j < vec[y].size(); j++) 165 { 166 int x = vec[y][j]; 167 sum -= (LL)oula[x]*(LL)(--cnt[x]); 168 } 169 xr--; 170 } 171 while(xr < ask[i].r) 172 { 173 xr++; 174 int y = ans[xr]; 175 for(j = 0; j < vec[y].size(); j++) 176 { 177 int x = vec[y][j]; 178 sum += (LL)oula[x]*(LL)(cnt[x]++); 179 } 180 } 181 answ[ask[i].id] = sum; 182 } 183 }
标签:iss number first query 欧拉 win main miss printf
原文地址:http://www.cnblogs.com/zzuli2sjy/p/6041033.html