Codeforces Round #273 (Div. 2)

In the first sample the following sequence of operations is possible:

1. One coin is passed from the fourth player to the second player;
2. One coin is passed from the fourth player to the fifth player;
3. One coin is passed from the first player to the third player;
4. One coin is passed from the fourth player to the second player.

分析: 签到，注意0的时候特判－1

#include <cstdio>
int main()
{
    int a, b, c, d, e;
    scanf("%d %d %d %d %d", &a, &b, &c, &d, &e);
    int sum = a + b + c + d + e;
    if(sum == 0)
        printf("-1\n");
    else if(sum % 5 == 0)
        printf("%d\n", sum / 5);
    else
        printf("-1\n");
}

5 1

10 10

3 2

1 1

6 3

3 6

分析: 贪心，最大值显然是n－m＋1人为一组，最小值则是尽可能均分，均分的话将余数均分分别加到商上即可

#include <cstdio>
#define ll long long

ll cal(ll x)
{
    return x * (x - 1) / 2;
}

int main()
{
    ll n, m;
    ll ma, mi;
    scanf("%I64d %I64d", &n, &m);
    ma = cal(n - m + 1);
    ll tmp = n % m;
    ll a = n / m;
    ll num = 0;
    if(tmp == 0)
        mi = m * cal(a);
    else
        mi = tmp * cal(a + 1) + (m - tmp) * cal(a);
    printf("%I64d %I64d\n", mi, ma);
}

5 4 3

4

1 1 1

1

2 3 3

2

分析:大概写几个yy一下

#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;

int main()
{
    ll a[3], ans = 0;
    scanf("%I64d %I64d %I64d", &a[0], &a[1], &a[2]);
    sort(a, a + 3);
    if((a[0] + a[1]) <= a[2] / 2)
        ans = a[0] + a[1];
    else
        ans = (a[0] + a[1] + a[2]) / 3;
    printf("%I64d\n",ans);
}

D. Red-Green Towers

time limit per test

2 seconds

memory limit per test

256 megabytes

There are r red andg green blocks for construction of the red-green tower. Red-green tower can be built following next rules:

• Red-green tower is consisting of some number of levels;

• Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n?-?1 blocks, the third one — ofn?-?2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;

• Each level of the red-green tower should contain blocks of the same color.

Let h be the maximum possible number of levels of red-green tower, that can be built out ofr red and g green blocks meeting the rules above. The task is to determine how many different red-green towers havingh levels can be built out of the available blocks.

Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.

You are to write a program that will find the number of different red-green towers of heighth modulo 10⁹?+?7.

Input

The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0?≤?r,?g?≤?2·10⁵,r?+?g?≥?1).

Output

Output the only integer — the number of different possible red-green towers of heighth modulo 10⁹?+?7.

Sample test(s)

Input

4 6

Output

2

Input

9 7

Output

6

Input

1 1

Output

2

Note

The image in the problem statement shows all possible red-green towers for the first sample.

#include <cstdio> 
#include <cstring>
#define ll long long   
int const MOD = 1e9 + 7;
int const MAX = 200000 + 5;
int dp[MAX];

int main()
{
    int r, g, h, ans = 0, tmp = 0;
    scanf("%d %d", &r, &g);
    ll min_r;
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    for(int i = 1; tmp <= (r + g) ;i++)
    {
        tmp += i;
        if(tmp > (r + g))
        {
            h = i - 1;
            break;
        }
    }
    min_r = h * (h + 1) / 2 - g;
    for(int i = 1; i <= h; i++) 
    {
        for(int j = r; j >= 0; j--)  
        {
            if(j - i >= 0)
                dp[j] = (dp[j] % MOD + dp[j - i] % MOD) % MOD;
            if(i == h && j >= min_r) 
                ans = (ans % MOD + dp[j] % MOD) % MOD;
        }
    }
    printf("%d\n", ans);
}