Crossings
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100463
Description
Input
There are several test cases in the input file. Each test case is specified by three space-separated numbers n, a, and b on a line. The prime n will be at most 1,000,000. The input is terminated with a line containing three zeros.
Output
For each case in the input print out the case number followed by the crossing number of the permutation. Follow the format in the example output.
Sample Input
5 2 1 19 12 7 0 0 0
Sample Output
Case 1: 3 Case 2: 77
HINT
思路:
很明显的逆序对。。照着公式把每个值算出来,然后直接求逆序对个数就好了。
注意要开long long 要不会超时。。。因为中间计算公式的时候值是超过了in范围t的,
不开long long就会陷入死循环
实现代码:
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #define ll long long using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define mid int m = (l + r) >> 1 const int M = 1e6 + 10; const double EPS = 1e-8; //inline int sgn(double x) return (x > EPS) - (x < -EPS); //浮点数比较常数优化写法 int c[M],l[M]; ll n; int lowbit(int x){ return x&(-x); } int getsum(int x){ int sum = 0; while(x>0){ sum += c[x]; x -= lowbit(x); } return sum; } void update(int x,int value){ while(x<=n){ c[x] += value; x += lowbit(x); } } int main() { ll a,b; int cal = 1; while(scanf("%lld%lld%lld",&n,&a,&b)!=EOF){ memset(c,0,sizeof(c)); if(n==0&&a==0&&b==0) break; for(int i = 1;i <= n;i ++){ l[i] = ((i-1)*a+b)%n+1; } ll ans = 0; for(int i = 1;i <= n;i ++){ update(l[i],1); ans +=i - getsum(l[i]); } printf("Case %d: %lld\n",cal++,ans); } return 0; }
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100463
Description
Input
There are several test cases in the input file. Each test case is specified by three space-separated numbers n, a, and b on a line. The prime n will be at most 1,000,000. The input is terminated with a line containing three zeros.
Output
For each case in the input print out the case number followed by the crossing number of the permutation. Follow the format in the example output.
Sample Input
5 2 1 19 12 7 0 0 0
Sample Output
Case 1: 3 Case 2: 77
HINT
思路:
很明显的逆序对。。照着公式把每个值算出来,然后直接求逆序对个数就好了。
注意要开long long 要不会超时。。。因为中间计算公式的时候值是超过了in范围t的,
不开long long就会陷入死循环
实现代码:
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #define ll long long using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define mid int m = (l + r) >> 1 const int M = 1e6 + 10; const double EPS = 1e-8; //inline int sgn(double x) return (x > EPS) - (x < -EPS); //浮点数比较常数优化写法 int c[M],l[M]; ll n; int lowbit(int x){ return x&(-x); } int getsum(int x){ int sum = 0; while(x>0){ sum += c[x]; x -= lowbit(x); } return sum; } void update(int x,int value){ while(x<=n){ c[x] += value; x += lowbit(x); } } int main() { ll a,b; int cal = 1; while(scanf("%lld%lld%lld",&n,&a,&b)!=EOF){ memset(c,0,sizeof(c)); if(n==0&&a==0&&b==0) break; for(int i = 1;i <= n;i ++){ l[i] = ((i-1)*a+b)%n+1; } ll ans = 0; for(int i = 1;i <= n;i ++){ update(l[i],1); ans +=i - getsum(l[i]); } printf("Case %d: %lld\n",cal++,ans); } return 0; }