UVa11401 - Triangle Counting
11401 - Triangle Counting
Problem G Triangle Counting
Input: Standard Input
Output: Standard Output
You are given n rods of length 1, 2…, n. You have to pick any 3 of them & build a triangle. How many distinct triangles can you make? Note that, two triangles will be considered different if they have at least 1 pair of arms with different length.
Input
The input for each case will have only a single positive integer n (3<=n<=1000000). The end of input will be indicated by a case with n<3. This case should not be processed.
Output
For each test case, print the number of distinct triangles you can make.
Sample Input Output for Sample Input
|
5 8 0 |
3 22 |
Problemsetter: Mohammad Mahmudur Rahman
1 法一！！ 2 /* 3 先打表！！！后推公式！！！ 4 5 i 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 6 7 f[i] 0 1 3 7 13 22 34 50 70 95 125 161 203 252 308 372 8 9 做差 1 2 4 6 9 12 16 20 25 30 36 42 49 56 64 10 11 再做差 1 2 2 3 3 4 4 5 5 6 6 7 7 8 12 13 在奇数项做差和偶数项做差即可！！！ 14 */ 15 16 17 /*//一开始TLE！！后来就把它用来打表了，推出了通项公式！！！ 18 19 记得一道做过fft的题，打这个表轻轻松松！！！ 20 #include<stdio.h> 21 #include<string.h> 22 #include<algorithm> 23 #include<math.h> 24 #include<queue> 25 #include<set> 26 #include<vector> 27 #include<bitset> 28 using namespace std; 29 typedef long long ll; 30 31 const int M=1000010; 32 ll s[M],f[M]; 33 int get(){ 34 char c; 35 int res=0; 36 while(c=getchar(),!isdigit(c)); 37 do{ 38 res=(res<<3)+(res<<1)+(c-'0'); 39 }while(c=getchar(),isdigit(c)); 40 return res; 41 } 42 43 */ 44 45 46 int main() 47 { 48 ll i,ans1,ans2,ans3,ans,n,m; 49 while(~scanf("%lld",&n)) 50 { 51 if(n<3)break; 52 ll k=0;f[n+1]=n; 53 while((++k)<=n)f[n+1-k]=f[n+1+k]=n-k; 54 f[1]=f[0]=0; 55 //for(i=0;i<=2*n;i++)printf("%lld**",f[i]); 56 for(i=1;i<=n;i++)f[i*2]--; 57 for(i=1;i<=2*n;i++)f[i]/=2; 58 //for(i=0;i<=2*n;i++)printf("%lld**",f[i]); 59 s[1]=0; 60 ll ans=0; 61 for(i=2;i<=2*n;i++)s[i]=s[i-1]+f[i]; 62 for(i=1;i<=n;i++) 63 { 64 ans+=s[2*n]-s[i]; 65 ans-=(i-1)*(n-i); 66 ans-=(n-1); 67 ans-=(n-i)*(n-i-1)/2; 68 } 69 printf("%lld**\n",ans); 70 } 71 return 0; 72 } 73 74 75 76 77 78 79 #include<stdio.h> 80 #include<string.h> 81 #include<algorithm> 82 #include<math.h> 83 #include<queue> 84 #include<set> 85 #include<vector> 86 #include<bitset> 87 using namespace std; 88 typedef long long ll; 89 90 int main() 91 { 92 ll ans,k,n; 93 while(1) 94 { 95 n=get(); 96 if(n<3)break; 97 if(n%2==0){k=n/2-1;ans=k*(4*k*k+3*k-1)/6;printf("%lld\n",ans);continue;} 98 k=(n-1)/2-1;ans=k*(4*k*k+3*k-1)/6+(n/2-1)*(n/2); 99 printf("%lld\n",ans); 100 } 101 return 0; 102 } 103 104 105 106 107 108 法2！！ 109 110 111 112 113 114 /* 115 递推： 116 设最大边为x的三角形有c[x]个，y+z>x； 117 118 y的值 z的个数 119 1 0 120 2 1 121 3 3 122 。。。。。 123 x-1 x-2 124 125 126 除去y=z的情况有（x-1）/2种！！ 127 128 剩下的再除以2即可！！ 129 130 C[x]= ((x-1)*(x-2)/2-(x-1)/2)/2; 131 132 */ 133 #include<stdio.h> 134 #include<string.h> 135 #include<algorithm> 136 #include<math.h> 137 #include<queue> 138 #include<set> 139 #include<vector> 140 #include<bitset> 141 using namespace std; 142 typedef long long ll; 143 144 const int M=1000010; 145 ll f[M]; 146 147 int get(){ 148 char c; 149 int res=0; 150 while(c=getchar(),!isdigit(c)); 151 do{ 152 res=(res<<3)+(res<<1)+(c-'0'); 153 }while(c=getchar(),isdigit(c)); 154 return res; 155 } 156 157 158 int main() 159 { 160 ll i; 161 int n; 162 f[3]=0; 163 for(i=4;i<M;i++)f[i]=f[i-1]+((i-1)*(i-2)/2-(i-1)/2)/2; 164 while(1) 165 { 166 n=get(); 167 if(n<3)break; 168 printf("%lld\n",f[n]); 169 } 170 return 0; 171 }
