把2019分成若干个素数相加，求有多少种分法。

元素完全相同的算同一种方法，比如2+2017=2019和2017+2=2019

以下是我的心路历程

这道五分的题我做了快一个小时，做出来还挺有成就感的_(:3」∠)_

按照惯例，能暴力先暴力，我飞快打出一个判断素数的函数，一个爆搜

果然2019太大了，搜不出来

我想可能是判断素数占了大量时间

于是我先把2019以下的素数先找了出来放到数组里

int prime[307]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,
127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,
271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,
439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,
613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,
797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,
983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,
1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,
1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,
1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,
1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,
1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,
1973,1979,1987,1993,1997,1999,2003,2011,2017}; 

还是搜不出来

我看着上面这十几行素数，突然想到了我学了一个月的dp，

以及yxc大佬讲的背包九讲（虽然我只学会了前几讲）

这么多素数不就是物品吗，2019不就是背包容积吗？

我只需要拿这些物品装满这个背包，求装满的方案数不就好了吗？

想到这儿，我赶紧把我的思路转换成代码(｀・ω・´)

    int i,j;
    int a[307][2020]={1};   //a[i][j]表示有前i个素数可用的情况下，能凑出j的方案数
    for(i=1;i<=306;i++)
    {
        for(j=0;j<=2019;j++)
        {
            a[i][j]=a[i-1][j];  //不用第i个素数的方案数
            if(j-prime[i]>=0)
                a[i][j]+=a[i-1][j-prime[i]];    //加上用上第i个素数的的方法数 
        }
    }

当然a[307][2020]的数组会爆掉

然后做一下空间优化就好了

最终代码如下

    long long a[2020]={1};
    for(int i=1;i<=306;i++)
    {
        for(int j=2019;j>=prime[i];j--)
            a[j]+=a[j-prime[i]];
    }
    printf("%lld",a[2019]);

答案为 55965365465060

每一个a[i][j]都能由它上一行左边的数(a[i-1][j-prime[i]])和它正上方的数推出来

内层循环从右向左遍历就可以保证对a[j]赋值时a[j-prime[i]]还是上一轮的值

本优化方法可以参考一维数组输出杨辉三角