hdu2082找单词(背包)

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找单词

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6398    Accepted Submission(s): 4478


Problem Description
假设有x1个字母A, x2个字母B,..... x26个字母Z,同时假设字母A的价值为1,字母B的价值为2,..... 字母Z的价值为26。那么,对于给定的字母,可以找到多少价值<=50的单词呢?单词的价值就是组成一个单词的所有字母的价值之和,比如,单词ACM的价值是1+3+14=18,单词HDU的价值是8+4+21=33。(组成的单词与排列顺序无关,比如ACM与CMA认为是同一个单词)。
 

Input
输入首先是一个整数N,代表测试实例的个数。
然后包括N行数据,每行包括26个<=20的整数x1,x2,.....x26.
 

Output
对于每个测试实例,请输出能找到的总价值<=50的单词数,每个实例的输出占一行。
 

Sample Input
21 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 09 2 6 2 10 2 2 5 6 1 0 2 7 0 2 2 7 5 10 6 10 2 10 6 1 9
 

Sample Output
7379297

刚看到这题以为是组合数学或者思维。。。一脸懵逼....

然后看了下discuss,是个背包,于是朝dp方向想了想1s,的确是个背包= =  (感觉自己贼菜啊

瞬间懂了。感觉这题还是很水的。。。
然后敲了敲背包就AC了= =
#include<cstdio>#include<cstdlib>#include<iostream>#include<stack>#include<queue>#include<algorithm>#include<string>#include<cstring>#include<cmath>#include<vector>#include<map>#include<set>#define eps 1e-8#define zero(x) (((x>0?(x):-(x))-eps)#define mem(a,b) memset(a,b,sizeof(a))#define memmax(a) memset(a,0x3f,sizeof(a))#define pfn printf("\n")#define ll __int64#define ull unsigned long long#define sf(a) scanf("%d",&a)#define sf64(a) scanf("%I64d",&a)#define sf264(a,b) scanf("%I64d%I64d",&a,&b)#define sf364(a,b,c) scanf("%I64d%I64d%I64d",&a,&b,&c)#define sf2(a,b) scanf("%d%d",&a,&b)#define sf3(a,b,c) scanf("%d%d%d",&a,&b,&c)#define sf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)#define sff(a) scanf("%f",&a)#define sfs(a) scanf("%s",a)#define sfs2(a,b) scanf("%s%s",a,b)#define sfs3(a,b,c) scanf("%s%s%s",a,b,c)#define sfd(a) scanf("%lf",&a)#define sfd2(a,b) scanf("%lf%lf",&a,&b)#define sfd3(a,b,c) scanf("%lf%lf%lf",&a,&b,&c)#define sfd4(a,b,c,d) scanf("%lf%lf%lf%lf",&a,&b,&c,&d)#define sfc(a) scanf("%c",&a)#define ull unsigned long long#define debug printf("***\n")const double PI = acos(-1.0);const double e = exp(1.0);const int INF = 0x7fffffff;;template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }template<class T> inline T Min(T a, T b) { return a < b ? a : b; }template<class T> inline T Max(T a, T b) { return a > b ? a : b; }bool cmpbig(int a, int b){ return a>b; }bool cmpsmall(int a, int b){ return a<b; }using namespace std;int main(){   // freopen("data.in","r",stdin);    int t;    sf(t);    while(t--)    {        int dp[30][60];        int num[30];        mem(dp,0);        for(int i=1;i<=26;i++)            sf(num[i]);        for(int i=0;i<=26;i++)            dp[i][0]=1;        for(int i=1;i<=26;i++)        {            for(int j=1;j<=50;j++)            {                dp[i][j]=dp[i-1][j];                for(int k=1;k<=num[i];k++)                    if(j-k*i>=0)                        dp[i][j]+=dp[i-1][j-k*i];            }        }        ll cnt=0;        for(int i=1;i<=50;i++)            cnt+=dp[26][i];        printf("%I64d\n",cnt);    }    return 0;}

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