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  • cf478D Red-Green Towers

    D. Red-Green Towers
    time limit per test
     2 seconds
    memory limit per test
     256 megabytes
    input
     standard input
    output
     standard output

    There are r red and g 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 — of n - 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 of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h 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 height h 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 height h 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.

    其实就是sb背包dp

    因为数组开的小了还wa了一次

     1 #include<cstdio>  
     2 #include<iostream>  
     3 #include<cstring>  
     4 #include<cstdlib>  
     5 #include<algorithm>  
     6 #include<cmath>  
     7 #include<queue>  
     8 #include<deque>  
     9 #include<set>  
    10 #include<map>  
    11 #include<ctime>  
    12 #define LL long long  
    13 #define inf 0x7ffffff  
    14 #define pa pair<int,int>  
    15 #define pi 3.1415926535897932384626433832795028841971  
    16 #define mod 1000000007  
    17 using namespace std;  
    18 inline LL read()  
    19 {  
    20     LL x=0,f=1;char ch=getchar();  
    21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}  
    22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}  
    23     return x*f;  
    24 }  
    25 LL a,b,h,ans;  
    26 LL f[1000010];  
    27 int main()  
    28 {  
    29     a=read();b=read();  
    30     for(int i=1;i<=1000;i++)  
    31       if (i*(i+1)/2<=a+b)h=i;  
    32     f[0]=1;  
    33     for (int i=1;i<=h;i++)  
    34     {  
    35       LL sum=i*(i+1)/2;  
    36       for (int j=b;j>=i;j--)  
    37         if (sum-j<=a)f[j]=(f[j]+f[j-i])%mod;  
    38     }  
    39     for (int i=0;i<=b;i++)  
    40       if (h*(h+1)/2-i<=a)ans=(ans+f[i])%mod;  
    41     printf("%lld
    ",ans);  
    42 }  
    cf478D
    ——by zhber,转载请注明来源
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  • 原文地址:https://www.cnblogs.com/zhber/p/4055138.html
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