写高数太累了,干点别的事情.
题目莫名其妙.
除了初始化,还有终止条件.
方案数问题,完全背包.这里第i个物品即第i个质数,需要得出刚好装满背包的方案总数.
先来一波打表把可能用到的素数(1~1000)搜索出来:
// 复制粘贴这个你就能成功WA了 #include <algorithm> #include <cstdio> #include <cstring> #include <iostream> using namespace std; bool isPrime(int n){ for(int i = 2; i * i <= n; i++) if(!(n % i)) return false; return true; } int main(){ for(int i = 2; i <= 1000; i++) if(isPrime(i)) cout << i << ", "; return 0; }
然后有递推式:dp[j] += dp[j - s[i]],s[i]为第i个素数(这里i从0开始,无关紧要).
s[i]是递增数列,循环条件写s[i] <= n.
初始化dp[0]为1,原因难以描述,但这样就处理好边界了.
加上背包模板:
// 复制粘贴这个你就又能成功WA了 #include <algorithm> #include <cstdio> #include <cstring> #include <iostream> using namespace std; int s[200] = {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}; int n, dp[1010]; int main(){ cin >> n; dp[0] = 1; for(int i = 0; s[i] <= n; i++) for(int j = s[i]; j <= n; j++) dp[j] += dp[j - s[i]]; cout << dp[n] << endl; return 0; }
不开long long见了祖宗.
// 复制粘贴这个就算开了long long你也会见祖宗 #include <algorithm> #include <cstdio> #include <cstring> #include <iostream> using namespace std; int s[200] = {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}; long long n, dp[1010]; int main(){ cin >> n; dp[0] = 1; for(int i = 0; s[i] <= n; i++) for(int j = s[i]; j <= n; j++) dp[j] += dp[j - s[i]]; cout << dp[n] << endl; return 0; }
该爆的还是会爆,因为n在998~1000的时候s[i] = 0,循环体没法跳出,然后就会裂开.
加一个素数就好了,以后记得打表多打一点.
#include <algorithm> #include <cstdio> #include <cstring> #include <iostream> using namespace std; int s[200] = {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}; long long n, dp[1010]; int main(){ cin >> n; dp[0] = 1; for(int i = 0; s[i] <= n; i++) for(int j = s[i]; j <= n; j++) dp[j] += dp[j - s[i]]; cout << dp[n] << endl; return 0; }