[POJ1320] Street Number

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题目描述

求解两个不相等的正整数n,m(n < m)使得1+2+3……+n=n+(n+1)+….+m

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输入格式

无输入

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输出格式

输出前10组满足条件的(n,m)(从小到大)，每组占一行。对于每组输出的n、m宽度为10。

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样例数据

样例输入

样例输出

前两组：
6 8
35 49

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题目分析

(2m+1)^2-8n^2=1
令x=2m+1，y=n,则有x^2-8y^2=1
典型佩尔方程，斗出最小解x1=3,y1=1
故由佩尔方程迭代公式得

xn=3xn−1+8yn−1

yn=xn−1+3yn−1

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源代码

#include<algorithm>#include<iostream>#include<iomanip>#include<cstring>#include<cstdlib>#include<vector>#include<cstdio>#include<cmath>#include<queue>using namespace std;inline const int Get_Int() {    int num=0,bj=1;    char x=getchar();    while(x<'0'||x>'9') {        if(x=='-')bj=-1;        x=getchar();    }    while(x>='0'&&x<='9') {        num=num*10+x-'0';        x=getchar();    }    return num*bj;}//n(n+1)/2=(m-n)(m+n+1)/2 即 (2m+1)^2-8n^2=1 //x(n+1)=3xn+8yn  y(n+1)=xn+3ynint Limit;int main() {    Limit=10;    int Lastx=3,Lasty=1;    for(int i=1; i<=Limit; i++) {        int x=3*Lastx+8*Lasty,y=Lastx+3*Lasty;        printf("%10d%10d\n",y,(x-1)/2);        Lastx=x;        Lasty=y;    }    return 0;}

也可以直接打表

#include<algorithm>#include<iostream>#include<iomanip>#include<cstring>#include<cstdlib>#include<vector>#include<cstdio>#include<cmath>#include<queue>using namespace std;inline const int Get_Int() {    int num=0,bj=1;    char x=getchar();    while(x<'0'||x>'9') {        if(x=='-')bj=-1;        x=getchar();    }    while(x>='0'&&x<='9') {        num=num*10+x-'0';        x=getchar();    }    return num*bj;}int main() {    printf("         6         8\n        35        49\n       204       288\n      1189      1681\n      6930      9800\n     40391     57121\n    235416    332928\n   1372105   1940449\n   7997214  11309768\n  46611179  65918161\n");    return 0;}

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