POJ-3132（一个数能表示成至多k个素数的和的种类数）（Sum of Different Primes ）

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又是动态规划

int ans[1125][15] = {0};int main(){int n, k, prim[1000], i, x = 0, len = 187, j;int d[] = {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};ans[0][0] = 1;for(i = 0; i < len; ++i)for (j = 1120 - d[i]; j >= 0; --j)for(k = 1; k <= 14; ++k)ans[j + d[i]][k] += ans[j][k - 1];while (scanf("%d%d", &n, &k) == 2 && (n || k))printf("%d\n", ans[n][k]);return 0;}