codeforces 317 A - Lengthening Sticks

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题目描述：

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.

题解：

很好地一道题目。给出以下解法。
（1）最笨最麻烦的解法：我们认定a是最的边,其他的<=a.那么我们暴力枚举a+x的x,设b+i,那么i的定义域我们知道了.对于一个i,c的可取范围其实是一个一次函数,但是这个一次函数是分段的.所以我们要耐心的解出一个有定义域且分段的一次函数.很麻烦,可以用很多if,也可以合起来写.之后是容斥的部分,很简单,暴力就好.
（2）思考一点反面：当我们认定a是最大的边的时候，b和c算反面。那么都有什么反面？b大于a，c大于a，b和c都小于等于a，但是b+c<=a. 前一部分用容斥,后一部分是重点:在三角形中很常用的技巧.b+c<=a,那么自然而然的b<=a,c<=a.这样就做了.
(3)整体思考反面.很好的思维:首先笨笨的算出所有的情况.之后减去不合法的:不合法的中,我们认定a是这种情况下最大的.那么,此时b和c已经天然有了<=a的性质,只需要减去b+c<=a的情况.之后我们容斥,a和b同样都是最大的,c比较小,这种情况没有不合法情况,abc都最大也没有不合法情况.所以就结束了.

重点：

(1)三角形判定,认定一条边是最大值.之后的只需要b+c>a就好了.
(2)解法一中的用函数表示对函数求和的方法.
(3)逆向思维.特别是三角形判定有一个最大边什么的.
(4)容斥的掌握,减去的东西应该是上一步大前提下多算的.

代码：

//三种解法都写在下面了#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <ctype.h>#include <limits.h>#include <cstdlib>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#include <set>#include <bitset>#define CLR(a) memset(a, 0, sizeof(a))#define REP(i, a, b) for(ll i = a;i < b;i++)#define REP_D(i, a, b) for(ll i = a;i <= b;i++)typedef long long ll;using namespace std;const double EPS = 1e-9;ll a, b, c, s;ll gao(ll a, ll b, ll xl, ll xr){    if(xl > xr)        return 0;    if(a*xl + b < 0 && a*xr+b<0)        return 0;    double key = -b*1.0/a;    ll keyx;    if(a > 0)    {        keyx = (key - EPS);        keyx++;    }    else    {        keyx = (key + EPS);    }    if(a*xl + b < 0)    {        xl = keyx;    }    if(a*xr + b < 0)    {        xr = keyx;    }    ll ans = b*(xr-xl+1);    ll tmp = (xr+xl)*(xr-xl+1)/2;    ans += tmp*a;    return ans;}ll first_old(ll a, ll b, ll c){    ll ans = 0;    ll i1, i2;    ll jl, jr;    ll tmp, oldAns;    for(ll x = 0; x<=s; x++)    {        tmp = 0;        oldAns = ans;        ll keyi = min(a+x-b, s - x);        if(keyi < 0)            continue;        i1 = a+x+1-b-c;        i2 = s-2*x-a+c;        i1 = min(keyi, i1);        i2 = min(keyi, i2);        if(i1 < 0 && i2 < 0)        {            jl = 0;            jr = keyi;            ans += gao(-1, s-x+1, jl, jr);        }        else if(i1 < 0)        {            jl = 0;            jr = i2;            ans += gao(0, a+x-c+1, jl, jr);            jl = i2+1;            jr = keyi;            ans += gao(-1, s-x+1, jl, jr);        }        else if(i2 < 0)        {            jl = 0;            jr = i1;            ans += gao(0, s - 2*x - a - 1 + b +c+1, jl, jr);            jl = i1+1;            jr = keyi;            ans += gao(-1, s-x+1, jl, jr);        }        else        {            if(i1==i2)            {                jl = 0;                jr = i1;                ans += gao(1, b-1+1, jl, jr);                jl = i1+1;                jr = keyi;                ans += gao(-1, s-x+1, jl, jr);            }            else if(i1 < i2)            {                jl = 0;                jr = i1;                ans += gao(1, b-1+1, jl, jr);                jl = i1+1;                jr = i2;                ans += gao(0, a+x-c+1, jl, jr);                jl = i2+1;                jr = keyi;                ans += gao(-1, s-x+1, jl, jr);            }            else            {                jl = 0;                jr = i2;                ans += gao(1, b-1+1, jl, jr);                jl = i2+1;                jr = i1;                ans += gao(0, s - 2*x - a - 1 + b +c+1, jl, jr);                jl = i1+1;                jr = keyi;                ans += gao(-1, s-x+1, jl, jr);            }        }//        for(ll bx = 0;bx+x<=s;bx++)//        {//            for(ll cx = 0;cx+bx+x<=s;cx++)//            {//                if(a+x >= b+bx && a+x >= c+cx)//                {//                    if(a+x+b+bx>c+cx && a+x+c+cx>b+bx && b+bx+c+cx>a+x)//                    {//                        tmp++;//                    }//                }//            }//        }        //printf("x is %I64d   ans is %I64d    tmp is %I64d\n", x, ans - oldAns, tmp);    }    return ans;}ll first(ll a, ll b, ll c){    ll ans = 0;    for(ll x = 0; x<=s; x++)    {        if(b > a+x || c > a+x)            continue;        ll tmp = (s - x + 2)*(s-x + 1)/2;        if(s-x-(a+x+1-b) >= 0)            tmp -= (s-x-(a+x+1-b)+2)*(s-x-(a+x+1-b)+1)/2;        if(s-x-(a+x+1-c) >= 0)            tmp -= (s-x-(a+x+1-c)+2)*(s-x-(a+x+1-c)+1)/2;        if(s - x - (a+x+1-b) - (a+x+1-c)>=0)        {            tmp += (s - x - (a+x+1-b) - (a+x+1-c)+2)*(s - x - (a+x+1-b) - (a+x+1-c)+1)/2;        }        ll sum = a+x - b - c;        if(sum >= 0)        {            sum = min(sum, s - x);            tmp -= (sum+2)*(sum+1)/2;        }        ans += tmp;//        ll tmp2 = 0;//        for(ll bx = 0; bx+x<=s; bx++)//        {//            for(ll cx = 0; cx+bx+x<=s; cx++)//            {//                if(a+x >= b+bx && a+x >= c+cx)//                {//                    if(a+x+b+bx>c+cx && a+x+c+cx>b+bx && b+bx+c+cx>a+x)//                    {//                        tmp2++;//                    }//                }//            }//        }//        printf("x is %I64d   ans is %I64d    tmp is %I64d\n", x, tmp, tmp2);    }    return ans;}ll second(ll a, ll b, ll c){    ll ans = 0;    for(ll i = max(a, b);; i++)    {        if(i < c)            continue;        ll t = i - a + i - b;        t = s - t;        if(t < 0)            break;        if(c + t > i)        {            ans += i-c+1;        }        else        {            ans += t+1;        }    }    return ans;}ll third(ll a, ll b, ll c){    ll ans = 0;    ll t = max(a, b);    t = max(t, c);    for(ll i = t;; i++)    {        ll used = 3*i - a - b - c;        if(used > s)            break;        ans++;    }    return ans;}void solve_old(){    ll ans = 0;    ans += first(a, b, c);    ans += first(b, a, c);    ans += first(c, a, b);    ans -= second(a, b, c);    ans -= second(b, c, a);    ans -= second(a, c, b);    ans += third(a, b, c);    printf("%I64d\n", ans);}ll new_first(ll a, ll b, ll c){    ll ans = 0;    for(ll x = 0;x<=s;x++)    {        if(a+x-b-c<0)            continue;        ll tmp = min(a+x-b-c, s-x);        ans += (tmp+2)*(tmp+1)/2;    }    return ans;}ll new_second(ll a, ll b, ll c){    ll ans = 0;    for(ll x = max(a, b);;x++)    {        if(x-a+x-b > s)            break;        ll tmp = s - (x-a+x-b);        if(2*x-c>0)            tmp -= (2*x-c);        if(tmp >= 0)            ans += tmp+1;    }    return ans;}void solve(){    ll ans = (s+3)*(s+2)*(s+1)/6;    ans -= new_first(a, b, c);    ans -= new_first(b, a, c);    ans -= new_first(c, a, b);    //ans += new_second(a, b, c);    //ans += new_second(b, c, a);    //ans += new_second(a, c, b);    printf("%I64d\n", ans);}int main(){    freopen("1Ain.txt", "r", stdin);    //freopen("1Aout.txt", "w", stdout);    while(scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &s) != EOF)    {        solve();    }    return 0;}