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题目大意：给定一个数n，求出一段连续的正整数的平方和等于n的方案数，并输出这些方案，注意输出格式；
循环判断条件可以适当剪支，提高效率，（1^2+2^2+..n^2)=n*(n+1)*(2n+1)/6;
尺取时一定要注意循环终止条件的判断。
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;#define rep(i,a,n) for (int i=a;i =a;i--)#define in(n) scanf("%d",&(n))#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))#define inll(n) scanf("%I64d",&(n))#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))#define inlld(n) scanf("%lld",&(n))#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))#define inf(n) scanf("%f",&(n))#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))#define inlf(n) scanf("%lf",&(n))#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))#define inc(str) scanf("%c",&(str))#define ins(str) scanf("%s",(str))#define out(x) printf("%d\n",(x))#define out2(x1,x2) printf("%d %d\n",(x1),(x2))#define outf(x) printf("%f\n",(x))#define outlf(x) printf("%lf\n",(x))#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));#define outll(x) printf("%I64d\n",(x))#define outlld(x) printf("%lld\n",(x))#define outc(str) printf("%c\n",(str))#define pb push_back#define mp make_pair#define fi first#define se second#define SZ(x) ((int)(x).size())#define mem(X,Y) memset(X,Y,sizeof(X));typedef vector vec;typedef long long ll;typedef pair P;const int dx[4]={ 1,0,-1,0},dy[4]={ 0,1,0,-1};const int INF=0x3f3f3f3f;const ll mod=1e9+7;ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;}const bool AC=true;struct node{ ll num,a,b;};node p[100000];bool cmp(node x,node y){ return x.num>y.num;}int main(){ ll n,s,t,sum,k,cnt;//都设为longlong免得溢出 while(inll(n)!=EOF){ s=t=1;sum=0;k=0; while(true){ while(sum n) break; //注意循环终止条件 } cnt=k; printf("%I64d\n",cnt);//别忘了这个 sort(p,p+k,cmp); rep(i,0,k){ printf("%I64d ",p[i].num); for(ll j=p[i].a;j<=p[i].b;j++){ printf("%I64d ",j); } printf("\n"); } } return 0;}
转载于:https://www.cnblogs.com/akrusher/p/5353125.html