【题目链接】:http://codeforces.com/contest/534/problem/B

【题意】

你在t秒内可以将车的速度任意增加减少绝对值不超过d;
然后要求在一开始车速为v1,t秒之后车速变为v2;
问你这段t时间内,车最多能行驶多远。

【题解】

枚举车“最大速度”v
看看车到达这个速度之后,然后回到速度v2(也就是说v是可能小于v2的,所以最大速度加了引号”)看看可不可行;
如果能在到达最大速度之后又回到速度v2(在t时间内);
那么记下回到v2的时间t1
在到达v和回到v2这段时间内的位移+(t-t1)*max(v,v2)就是答案了

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)typedef pair<int, int> pii;
typedef pair<LL, LL> pll;const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 110;int v1, v2, t, d,ans = 0;int main()
{//freopen("F:\\rush.txt", "r", stdin);rei(v1), rei(v2), rei(t), rei(d);rep1(v, v1, v1 + t*d){int temp = 0;int now = 0, vv = v1;while (vv < v){temp += vv;vv += d;vv = min(vv, v);now++;if (now > t)break;}if (now > t) continue;if (vv < v2){while (vv < v2){temp += vv;vv += d;vv = min(vv, v2);now++;if (now > t)break;}if (now > t)continue;now++;temp += vv;}elseif (vv > v2){while (vv > v2){temp += vv;vv -= d;vv = max(vv, v2);now++;if (now > t)break;}if (now > t)continue;now++;temp += vv;}elseif (vv == v2){now++;temp += vv;}temp += max(v, vv)*(t - now);ans = max(ans, temp);}printf("%d\n", ans);//printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);return 0;
}