分段打表

适用题目特征

当打表范围过大，但表之间可以递推。

原理

借助分块的思想，答案来自若干子表的组合。即，整块的答案用预处理的值计算，非整块的答案暴力计算。
关于分块大小，通常会取得最好的效果，但受限于提交代码长度限制，实际分块大小通常大于。
PS：2020年CCPC网络赛中，计算的素数和的题目，似乎可以用这种方法水过去。

例题

Luogu P1822
代码如下

/*
Luogu P1822
*/
#define method_1
#ifdef method_1
/*

*/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define D(x) cout<<#x<<" = "<=(s);i--)
using namespace std;
typedef long long ll;
typedef pairpii;
const int maxn=1000+5;
const int Len=1000000; //sqrt(1e9)
const int INF=0x3f3f3f3f;
const int cnt[]={0,2045,1044,750,357,316,316,358,749,1044,1474,866,1044,303,220,154,115,84,125,346,927,378,558,750,221,204,154,118,134,146,382,110,270,303,357,176,204,168,128,89,75,85,126,298,221,316,176,220,168,88,69,69,88,168,220,176,316,221,298,126,85,75,89,128,168,204,176,357,303,270,111,382,145,134,118,154,204,222,749,558,378,927,347,124,84,115,154,220,303,1044,866,2022,754,224,71,86,115,168,298,558,1474,626,558,118,81,51,33,19,30,316,1461,866,1044,303,220,154,115,84,125,346,927,110,558,94,221,96,77,48,32,26,60,74,270,116,66,95,204,118,83,55,29,50,126,118,77,50,76,113,168,57,43,44,88,108,79,40,41,44,91,98,85,41,89,69,81,34,23,30,21,80,67,136,145,74,64,51,21,22,24,35,53,489,347,75,61,51,24,13,28,172,79,1143,754,47,45,51,36,15,20,229,924,240,240,298,118,69,51,28,43,63,585,866,328,303,113,77,51,35,25,62,375,378,558,750,221,204,154,118,134,146,382,67,194,303,84,176,96,81,49,32,25,47,98,298,126,69,95,220,108,52,38,43,57,168,113,76,50,77,118,126,50,48,69,128,118,60,40,34,29,96,74,156,97,134,76,77,34,27,23,26,55,162,193,124,54,69,51,23,23,35,37,311,315,224,35,58,51,22,28,72,277,25,78,118,168,65,58,35,27,14,14,382,92,116,220,101,69,53,33,39,32,110,558,94,221,96,77,48,32,26,60,110,270,303,357,176,204,168,128,89,75,34,83,118,221,64,176,113,109,42,27,38,52,108,220,95,69,126,298,98,47,35,54,69,168,96,76,59,65,194,110,51,66,74,118,101,60,44,30,47,138,32,60,75,84,65,77,52,21,25,33,24,41,47,71,52,69,56,34,35,16,23,47,91,118,115,52,45,35,16,15,170,64,65,113,154,65,61,32,32,20,378,240,79,126,204,101,64,46,27,54,67,194,303,84,176,96,81,49,32,25,85,126,298,221,316,176,220,168,88,69,27,42,109,113,176,64,221,118,83,34,29,55,83,118,204,95,66,116,270,74,41,49,78,76,154,96,77,33,78,110,23,46,49,54,115,101,79,29,23,52,15,22,53,35,86,65,81,52,26,12,12,26,52,81,65,86,35,53,22,15,52,23,29,79,101,115,54,49,46,23,110,78,33,77,96,154,76,78,49,41,74,270,116,66,95,204,118,83,55,29,34,83,118,221,64,176,113,109,42,27,69,88,168,220,176,316,221,298,126,85,25,32,49,81,96,176,84,303,194,67,54,27,46,64,101,204,126,79,240,378,20,32,32,61,65,154,113,65,64,170,15,16,35,45,52,115,118,91,47,23,16,35,34,56,69,52,71,47,41,24,33,25,21,52,77,65,84,75,60,32,138,47,30,44,60,101,118,74,66,51,110,194,65,59,76,96,168,69,54,35,47,98,298,126,69,95,220,108,52,38,27,42,109,113,176,64,221,118,83,34,75,89,128,168,204,176,357,303,270,110,60,26,32,48,77,96,221,94,558,110,32,39,33,53,69,101,220,116,92,382,14,14,27,35,58,65,168,118,78,26,276,72,28,22,51,58,36,224,315,310,37,35,23,23,51,69,54,125,193,161,55,26,23,27,34,77,76,134,98,155,74,96,29,34,40,60,118,128,69,48,50,126,118,77,50,76,113,168,57,43,38,52,108,220,95,69,126,298,98,47,25,32,49,81,96,176,84,303,194,68,382,145,134,118,154,204,222,749,558,379,374,62,25,35,51,77,113,304,327,867,584,63,43,28,51,69,118,298,241,239,924,229,20,15,36,51,45,47,754,1143,79,172,28,13,24,51,61,76,346,489,53,35,24,22,21,51,64,74,145,136,67,80,21,30,23,34,81,69,89,41,85,98,91,44,41,40,79,108,88,44,43,57,168,113,76,50,77,118,126,50,29,55,83,118,204,95,66,116,270,74,60,26,32,48,77,96,221,94,558,110,927,347,124,84,115,154,220,303,1044,866,1461,316,30,19,33,51,81,118,558,626,3567,2257,88,6,20,33,35,54,314,2022,371,1006,134,9,9,36,53,50,192,927,57,72,84,14,11,24,48,78,98,382,46,41,23,18,9,21,56,83,69,75,50,83,52,27,21,23,52,109,57,69,44,88,108,79,40,41,44,91,98,85,35,54,69,168,96,76,59,65,194,110,54,27,46,64,101,204,126,79,240,379,374,62,25,35,51,77,113,304,327,866,2022,754,224,71,86,115,168,298,558,1474};
int A,B;
int ID(int x){
    return (x+Len-1)/Len;
}
int L(int x){
    return (ID(x)-1)*Len+1;
}
int R(int x){
    return ID(x)*Len;
}
inline int qpow(int a,int b) {
    int res=1;
    while(b){
        if(b&1) res=res*a;
        a=a*a;
        b>>=1;
    }
    return res;
}
int magic(int x) {
    int y=0,num[10],i=-1;
    while(x) num[++i]=x%10,x/=10;
    for(int j=0;j=7) res=1;
    if(ID(A)==ID(B)) rep(i,A,B) res+=magic(i)==7;
    else{
        rep(i,A,R(A)) res+=magic(i)==7;
        rep(i,L(B),B) res+=magic(i)==7;
        rep0(i,ID(A)+1,ID(B)) res+=cnt[i];
    }
    cout<>A>>B;
    solve();
    return 0;
}
#endif
#ifdef method_2
/*

*/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define D(x) cout<<#x<<" = "<=(s);i--)
using namespace std;
typedef long long ll;
typedef pairpii;
const int maxn=+5;
const int INF=0x3f3f3f3f;
inline int qpow(int a,int b) {
    int res=1;
    while(b){
        if(b&1) res=res*a;
        a=a*a;
        b>>=1;
    }
    return res;
}
int magic(int x) {
    int y=0,num[10],i=-1;
    while(x) num[++i]=x%10,x/=10;
    for(int j=0;j

分段打表

适用题目特征

原理

例题

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