codeforces #317 C. Lengthening Sticks (很好的想法题)

题目：http://codeforces.com/problemset/problem/572/C

题意：给定长为a,b,c的3个木棍，和一个值为l的buf。求使得(a+x,b+y,c+z)构成一个三角形的方案数，其中x+y+z<=l,x>=0,y>=0,z>=0。

C. Lengthening Sticks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given three sticks with positive integer lengths of a, b, and c centimeters. You can increase length of some of them by some positive integer number of centimeters (different sticks can be increased by a different length), but in total by at most l centimeters. In particular, it is allowed not to increase the length of any stick.

Determine the number of ways to increase the lengths of some sticks so that you can form from them a non-degenerate (that is, having a positive area) triangle. Two ways are considered different, if the length of some stick is increased by different number of centimeters in them.

Input

The single line contains 4 integers a, b, c, l (1 ≤ a, b, c ≤ 3·105, 0 ≤ l ≤ 3·105).

Output

Print a single integer — the number of ways to increase the sizes of the sticks by the total of at most l centimeters, so that you can make a non-degenerate triangle from it.

Sample test(s)

input

1 1 1 2

output

4

input

1 2 3 1

output

2

input

10 2 1 7

output

0

Note

In the first sample test you can either not increase any stick or increase any two sticks by 1 centimeter.

In the second sample test you can increase either the first or the second stick by one centimeter. Note that the triangle made from the initial sticks is degenerate and thus, doesn't meet the conditions.

分析：可以构成三角形的方案数=总方案数-不能构成三角形的方案数。

总的方案数N，就是从l中取出x点长度（0<=x<=l），将x分给a,b,c。怎么算将x分给a,b,c的方案数？

其实就是一个可重复选的组合问题，答案就是C(X+2,2)。

//ps:证明参考刘汝佳的训练指南p105,不过刘汝佳上面的证明有一点错误，已改正。

简单证明一下：有n个不同元素,每个元素可选多次,求一种选k次的方案数。假设第i个元素选xi次,

那么x1+x2+...+xn=k,其中xi>=0。令yi=xi+1,那么y1+y2+...+yn=k+n,其中yi>=1。那么问题就可以转变为有k+n个数字"1"排成一排，将这k+n个"1"分成n部分的方案数，由于每一部分的个数>=1，所以就是在k+n-1个分割点选出n-1个分割点。即方案数为C(k+n-1,n-1)=C(n+k-1,k)。

不能构成三角形的方案数。选a,b,c，中的其中一条边当最长边。枚举l分给最长边的点数就行。

代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;typedef unsigned long long ULL;const LL INF = 1E9+9;const int maxn = 1e6;LL sum[maxn];void Init(){sum[0]=1;for(int i=1;i<maxn;i++)sum[i]=i+1+sum[i-1];}inline LL cal(long long n){return (2+n)*(1+n)/2;}LL fun(LL a,LL b,LL c,LL n){LL ret(0);for(LL i=0;i<=n;i++){LL cur=a+i;if(cur<b+c)continue ;LL temp=min(cur-(b+c),n-i);ret+=sum[temp];}return ret;}int main(){Init();int a,b,c,l,i,j;LL ans=0;cin>>a>>b>>c>>l;for(i=0;i<=l;i++)ans+=cal(i);ans-=fun(a,b,c,l);ans-=fun(b,a,c,l);ans-=fun(c,a,b,l);cout<<ans<<endl;return 0;}