https://www.luogu.org/problemnew/show/P1217

考慮暴力生成所有的回文數(shù)然后再判斷是不是質(zhì)數(shù)。注意個位的選擇實際上只有4種。所以是 $4*10^3*10^3=4*10^6$ ，完全充裕的復(fù)雜度。

#include<bits/stdc++.h> using namespace std; #define ll long longint a,b; vector<int> ans;bool isp(int n){int c=sqrt(n+1e-4);for(int i=2;i<=c;i++){if(n%i==0)return false;}return true; }void gen1(){if(a<=2&&2<=b)ans.push_back(2);if(a<=3&&3<=b)ans.push_back(3);if(a<=5&&5<=b)ans.push_back(5);if(a<=7&&7<=b)ans.push_back(7); }void gen2(){for(int d=1;d<=9;d++){if(d!=5){int p=11*d;if(a<=p&&p<=b)if(isp(p))ans.push_back(p);}} }void gen3(){for(int d=1;d<=9;d++){if(d!=5){for(int d2=0;d2<=9;d2++){int p=101*d+10*d2;if(a<=p&&p<=b)if(isp(p))ans.push_back(p);}}} }void gen4(){for(int d=1;d<=9;d++){if(d!=5){for(int d2=0;d2<=9;d2++){int p=1001*d+110*d2;if(a<=p&&p<=b)if(isp(p))ans.push_back(p);}}} }void gen5(){for(int d=1;d<=9;d++){if(d!=5){for(int d2=0;d2<=9;d2++){for(int d3=0;d3<=9;d3++){int p=10001*d+1010*d2+100*d3;if(a<=p&&p<=b)if(isp(p))ans.push_back(p);}}}} }void gen6(){for(int d=1;d<=9;d++){if(d!=5){for(int d2=0;d2<=9;d2++){for(int d3=0;d3<=9;d3++){int p=100001*d+10010*d2+1100*d3;if(a<=p&&p<=b)if(isp(p))ans.push_back(p);}}}} }void gen7(){for(int d=1;d<=9;d++){if(d!=5){for(int d2=0;d2<=9;d2++){for(int d3=0;d3<=9;d3++){for(int d4=0;d4<=9;d4++){int p=1000001*d+100010*d2+10100*d3+1000*d4;if(a<=p&&p<=b)if(isp(p))ans.push_back(p);}}}}} }void gen8(){for(int d=1;d<=9;d++){if(d!=5){for(int d2=0;d2<=9;d2++){for(int d3=0;d3<=9;d3++){for(int d4=0;d4<=9;d4++){int p=10000001*d+1000010*d2+100100*d3+11000*d4;if(a<=p&&p<=b)if(isp(p))ans.push_back(p);}}}}} }void solve(){gen1();gen2();gen3();gen4();gen5();gen6();gen7();gen8(); }int main(){scanf("%d%d",&a,&b);solve();sort(ans.begin(),ans.end());for(auto i:ans)printf("%d\n",i); }

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轉(zhuǎn)載于:https://www.cnblogs.com/Yinku/p/10315468.html

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