CodeForces 18B Platforms (水题)

Description
In one one-dimensional world there are n platforms. Platform with index k (platforms are numbered from 1) is a segment with coordinates [(k - 1)m, (k - 1)m + l], and l < m. Grasshopper Bob starts to jump along the platforms from point 0, with each jump he moves exactly d units right. Find out the coordinate of the point, where Bob will fall down. The grasshopper falls down, if he finds himself not on the platform, but if he finds himself on the edge of the platform, he doesn't fall down.
Input
The first input line contains 4 integer numbers n, d, m, l (1 ≤ n, d, m, l ≤ 106, l < m) — respectively: amount of platforms, length of the grasshopper Bob's jump, and numbers m and l needed to find coordinates of the k-th platform: [(k - 1)m, (k - 1)m + l].
Output
Output the coordinates of the point, where the grosshopper will fall down. Don't forget that if Bob finds himself on the platform edge, he doesn't fall down.
Sample Input
Input
2 2 5 3
Output
4
Input
5 4 11 8
Output
20
#include<cmath>#include<queue>#include<stack>#include<vector>#include<cstdio>#include<cstring>#include<string>#include<map>#include<iostream>#include<algorithm>#include<functional>using namespace std;typedef long long LL;struct node{    long long l,r;}p[1000005];int n,d,m,l;int main(){    int i,j,k,T;    while(scanf("%d%d%d%d",&n,&d,&m,&l)!=EOF)    {        p[1].l=0;p[1].r=l;        for(i=2;i<=n;i++)         p[i].l=p[i-1].l+m,p[i].r=p[i-1].r+m;        long long now=0,ans;        p[n+1].l=1000000000;        p[n+1].l*=p[n+1].l;        for(i=1;i<=n;i++)        {            long long r=p[i].r;            long long x=r-now;            x%=d;            now=r-x;            now+=d;            if(now<p[i+1].l) {ans=now;break;}            while(now>p[i+1].r&&i<n) i++;            if(now<p[i+1].l) {ans=now;break;}        }        ans=now;        cout<<ans<<endl;    }    return 0;}