C 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2021-10-14

#include <limits.h>
int main(){
    int n;
    scanf("%d",&n);
    char**d=(char**)malloc(n*sizeof(char*));
    d[0]=(char*)malloc(50*sizeof(char));
    d[0]="1";
    d[1]=(char*)malloc(50*sizeof(char));
    d[1]="1";
    
    for(int i=2;i<=n;i++){
        int c=0;
        d[i]=(char*)malloc(50*sizeof(char));
        int l1=strlen(d[i-1]);
        int l2=strlen(d[i-2]);
       // printf("%d      %d\n",l1,l2);
        int l=0;
        while(l<l2){
            int x=d[i-1][l]-'0'+d[i-2][l]-'0'+c;
            if(x>=10){
                c=1;
                x=x%10;
            }
            else{
                c=0;
            }
            d[i][l++]=x+'0';
        }
        while(l<l1){
            int x=d[i-1][l]-'0'+c;
            if(x>=10){
                c=1;
                x=x%10;
            }
            else{
                c=0;
            }
            d[i][l++]=x+'0';
        }
        if(c==1){
            d[i][l++]='1';
        }
        d[i][l]='\0';
       // printf("%s\n",d[i]);
    }
    int l=strlen(d[n]);
    for(int i=l-1;i>=0;i--){
        printf("%c",d[n][i]);
    }
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 376KB, 提交时间: 2020-08-29

#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

int a[500][100] = { 0 };

int main()
{
	int n, i, j, r, k = 0, t, p = 0;
	scanf("%d", &n);
	a[0][99] = 1, a[1][99] = 1, a[2][99] = 2;//, a[3][99] = 2;
	for (j = 3; j <= n; ++j)
	{
		for (r = 99; r >= 0; --r)
		{
			t = a[j - 1][r] + a[j - 2][r] + p;
			if (t >= 10)
			{
				p = t / 10;
				a[j][r] = t % 10;
			}
			else
			{
				a[j][r] = t;
				p = 0;
			}
		}
	}
	for (j = 0; j <= 99; ++j)
	{
		if (k || a[n][j])
		{
			k = 1;
			printf("%d", a[n][j]);
		}
	}

}

C 解法, 执行用时: 2ms, 内存消耗: 376KB, 提交时间: 2020-08-22

#include <stdio.h>
   
int a[500][100];
   
int main()
{
    int n, i, j, r, k = 0, t, p = 0;
    scanf("%d", &n);
    a[0][99] = 1, a[1][99] = 1, a[2][99] = 2;//, a[3][99] = 2;
    for(j = 3; j <= n; ++j)
    {
        for(r = 99; r >= 0; --r)
        {
            t = a[j-1][r] + a[j-2][r] + p;
            if(t >= 10)
            {
                p = t/10;
                a[j][r] = t%10;
            }
            else
            {
                a[j][r] = t;
                p = 0;
             }
        }
    }
    for(j = 0; j <= 99; ++j)
    {
        if(k || a[n][j])
        {
            k = 1;
            printf("%d", a[n][j]);
        }
    }
 
}

C 解法, 执行用时: 2ms, 内存消耗: 376KB, 提交时间: 2020-08-21

#include<stdio.h>
#define MAX 101

typedef unsigned long long uint64;

#define maxdll (uint64)1e19

int ansh[MAX] = {0};
uint64 ansl[MAX] = {1,1};

void ullAdd (uint64* ansl, int* ansh, int i)
{
    ansl[i] = ansl[i-2]+ansl[i-1];
    ansh[i] = ansh[i-2]+ansh[i-1];
    if (ansl[i]>=maxdll)
    {
        ansh[i] ++;
        ansl[i] = ansl[i]-maxdll;
    }
}

int main()
{
    for(int i=1;i<MAX;i++)
    {
        ansh[i] = 0;
    }
    int n = 0;
    scanf("%d",&n);
    for (int i=2;i<=n;i++)
        ullAdd(ansl,ansh,i);
    if (ansh[n]>0)
        printf("%llu%llu",ansh[n],ansl[n]);
    else
        printf("%llu",ansl[n]);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 376KB, 提交时间: 2020-08-19

#include <stdio.h>
   
int a[500][100];
   
int main()
{
    int n, i, j, r, k = 0, t, p = 0;
    scanf("%d", &n);
    a[0][99] = 1, a[1][99] = 1, a[2][99] = 2;//, a[3][99] = 2;
    for(j = 3; j <= n; ++j)
    {
        for(r = 99; r >= 0; --r)
        {
            t = a[j-1][r] + a[j-2][r] + p;
            if(t >= 10)
            {
                p = t/10;
                a[j][r] = t%10;
            }
            else
            {
                a[j][r] = t;
                p = 0;
             }
        }
    }
    for(j = 0; j <= 99; ++j)
    {
        if(k || a[n][j])
        {
            k = 1;
            printf("%d", a[n][j]);
        }
    }
 
}