【循环】洛谷题目记录

2024-01-01 20:21:03

我比较菜，只是想单纯记录一下学习过程QAQ

欢迎大佬们指出错误以及更好的方案QWQ

质因数分解

质因数分解

思路：输入一定为两个质数的乘积，所以输入的整数n一定只有1和它本身以及两个不同质数这四个因数，利用这个特点便可以知道：从2开始循环寻找这两个质数，只需要找到其中一个便可以退出循环。

误区：从2开始循环每个数字并判断其是否为质数一定会超时。

由此，写出代码如下：

#include<stdio.h>
#include<math.h>
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 2;i<=sqrt(n);i++){
        if(n%i==0){
            printf("%d",n/i);
            break;
        }
    }
    return 0;
}

PS：可以确定较大的质数一定小于sqrt(n)，但是因为找到一个便可以结束循环，所以循环条件为i<n也可以。

回文质数

思路：输入为一个闭区间，目的是输出回文的质数。只需要确定一个数字是回文并且为质数就可以输出该数字。但是鉴于最大不会超过一亿，并且回文质数个数有限以及时间要求较高，可以考虑暴力打表。

误区：从左端点开始循环到右端点逐个判断一定会超时。因为每次都要用循环从2到sqrt(n)判断是不是质数，同时还需要循环判断是不是回文的，循环次数过多。

代码如下：

#include<stdio.h>

int main()
{
    int c[800]={5,7,11,101,131,151,181,
191,313,353,373,383,727,757,787,797,
919,929,10301,10501,10601,11311,11411,12421,12721,
12821,13331,13831,13931,14341,14741,15451,15551,16061,
16361,16561,16661,17471,17971,18181,18481,19391,19891,
19991,30103,30203,30403,30703,30803,31013,31513,32323,
32423,33533,34543,34843,35053,35153,35353,35753,36263,
36563,37273,37573,38083,38183,38783,39293,70207,70507,
70607,71317,71917,72227,72727,73037,73237,73637,74047,
74747,75557,76367,76667,77377,77477,77977,78487,78787,
78887,79397,79697,79997,90709,91019,93139,93239,93739,
94049,94349,94649,94849,94949,95959,96269,96469,96769,
97379,97579,97879,98389,98689,1003001,1008001,1022201,1028201,
1035301,1043401,1055501,1062601,1065601,1074701,1082801,1085801,1092901,
1093901,1114111,1117111,1120211,1123211,1126211,1129211,1134311,1145411,
1150511,1153511,1160611,1163611,1175711,1177711,1178711,1180811,1183811,
1186811,1190911,1193911,1196911,1201021,1208021,1212121,1215121,1218121,
1221221,1235321,1242421,1243421,1245421,1250521,1253521,1257521,1262621,
1268621,1273721,1276721,1278721,1280821,1281821,1286821,1287821,1300031,
1303031,1311131,1317131,1327231,1328231,1333331,1335331,1338331,1343431,
1360631,1362631,1363631,1371731,1374731,1390931,1407041,1409041,1411141,
1412141,1422241,1437341,1444441,1447441,1452541,1456541,1461641,1463641,
1464641,1469641,1486841,1489841,1490941,1496941,1508051,1513151,1520251,
1532351,1535351,1542451,1548451,1550551,1551551,1556551,1557551,1565651,
1572751,1579751,1580851,1583851,1589851,1594951,1597951,1598951,1600061,
1609061,1611161,1616161,1628261,1630361,1633361,1640461,1643461,1646461,
1654561,1657561,1658561,1660661,1670761,1684861,1685861,1688861,1695961,
1703071,1707071,1712171,1714171,1730371,1734371,1737371,1748471,1755571,
1761671,1764671,1777771,1793971,1802081,1805081,1820281,1823281,1824281,
1826281,1829281,1831381,1832381,1842481,1851581,1853581,1856581,1865681,
1876781,1878781,1879781,1880881,1881881,1883881,1884881,1895981,1903091,
1908091,1909091,1917191,1924291,1930391,1936391,1941491,1951591,1952591,
1957591,1958591,1963691,1968691,1969691,1970791,1976791,1981891,1982891,
1984891,1987891,1988891,1993991,1995991,1998991,3001003,3002003,3007003,
3016103,3026203,3064603,3065603,3072703,3073703,3075703,3083803,3089803,
3091903,3095903,3103013,3106013,3127213,3135313,3140413,3155513,3158513,
3160613,3166613,3181813,3187813,3193913,3196913,3198913,3211123,3212123,
3218123,3222223,3223223,3228223,3233323,3236323,3241423,3245423,3252523,
3256523,3258523,3260623,3267623,3272723,3283823,3285823,3286823,3288823,
3291923,3293923,3304033,3305033,3307033,3310133,3315133,3319133,3321233,
3329233,3331333,3337333,3343433,3353533,3362633,3364633,3365633,3368633,
3380833,3391933,3392933,3400043,3411143,3417143,3424243,3425243,3427243,
3439343,3441443,3443443,3444443,3447443,3449443,3452543,3460643,3466643,
3470743,3479743,3485843,3487843,3503053,3515153,3517153,3528253,3541453,
3553553,3558553,3563653,3569653,3586853,3589853,3590953,3591953,3594953,
3601063,3607063,3618163,3621263,3627263,3635363,3643463,3646463,3670763,
3673763,3680863,3689863,3698963,3708073,3709073,3716173,3717173,3721273,
3722273,3728273,3732373,3743473,3746473,3762673,3763673,3765673,3768673,
3769673,3773773,3774773,3781873,3784873,3792973,3793973,3799973,3804083,
3806083,3812183,3814183,3826283,3829283,3836383,3842483,3853583,3858583,
3863683,3864683,3867683,3869683,3871783,3878783,3893983,3899983,3913193,
3916193,3918193,3924293,3927293,3931393,3938393,3942493,3946493,3948493,
3964693,3970793,3983893,3991993,3994993,3997993,3998993,7014107,7035307,
7036307,7041407,7046407,7057507,7065607,7069607,7073707,7079707,7082807,
7084807,7087807,7093907,7096907,7100017,7114117,7115117,7118117,7129217,
7134317,7136317,7141417,7145417,7155517,7156517,7158517,7159517,7177717,
7190917,7194917,7215127,7226227,7246427,7249427,7250527,7256527,7257527,
7261627,7267627,7276727,7278727,7291927,7300037,7302037,7310137,7314137,
7324237,7327237,7347437,7352537,7354537,7362637,7365637,7381837,7388837,
7392937,7401047,7403047,7409047,7415147,7434347,7436347,7439347,7452547,
7461647,7466647,7472747,7475747,7485847,7486847,7489847,7493947,7507057,
7508057,7518157,7519157,7521257,7527257,7540457,7562657,7564657,7576757,
7586857,7592957,7594957,7600067,7611167,7619167,7622267,7630367,7632367,
7644467,7654567,7662667,7665667,7666667,7668667,7669667,7674767,7681867,
7690967,7693967,7696967,7715177,7718177,7722277,7729277,7733377,7742477,
7747477,7750577,7758577,7764677,7772777,7774777,7778777,7782877,7783877,
7791977,7794977,7807087,7819187,7820287,7821287,7831387,7832387,7838387,
7843487,7850587,7856587,7865687,7867687,7868687,7873787,7884887,7891987,
7897987,7913197,7916197,7930397,7933397,7935397,7938397,7941497,7943497,
7949497,7957597,7958597,7960697,7977797,7984897,7985897,7987897,7996997,
9002009,9015109,9024209,9037309,9042409,9043409,9045409,9046409,9049409,
9067609,9073709,9076709,9078709,9091909,9095909,9103019,9109019,9110119,
9127219,9128219,9136319,9149419,9169619,9173719,9174719,9179719,9185819,
9196919,9199919,9200029,9209029,9212129,9217129,9222229,9223229,9230329,
9231329,9255529,9269629,9271729,9277729,9280829,9286829,9289829,9318139,
9320239,9324239,9329239,9332339,9338339,9351539,9357539,9375739,9384839,
9397939,9400049,9414149,9419149,9433349,9439349,9440449,9446449,9451549,
9470749,9477749,9492949,9493949,9495949,9504059,9514159,9526259,9529259,
9547459,9556559,9558559,9561659,9577759,9583859,9585859,9586859,9601069,
9602069,9604069,9610169,9620269,9624269,9626269,9632369,9634369,9645469,
9650569,9657569,9670769,9686869,9700079,9709079,9711179,9714179,9724279,
9727279,9732379,9733379,9743479,9749479,9752579,9754579,9758579,9762679,
9770779,9776779,9779779,9781879,9782879,9787879,9788879,9795979,9801089,
9807089,9809089,9817189,9818189,9820289,9822289,9836389,9837389,9845489,
9852589,9871789,9888889,9889889,9896989,9902099,9907099,9908099,9916199,
9918199,9919199,9921299,9923299,9926299,9927299,9931399,9932399,9935399,
9938399,9957599,9965699,9978799,9980899,9981899,9989899};
    int a,b;
    scanf("%d%d",&a,&b);
    for(int i = 0;i<779;i++){
        if(c[i]>=a&&c[i]<=b)    printf("%d\n",c[i]);
    }
    return 0;
}

PS：如果不考虑时间限制，可以i按照如下思路

#include<stdio.h>
#include<math.h>
int judge(int a){
    int i,k = a,n = a,sum = 0;
    //回文
    for(i=0;k!=0;i++){
        k/=10;
    }
    k = i;
    for(i=0;i<k;i++){
        sum*=10;
        sum+=n%10;
        n/=10;
    }
    if(a!=sum)  return 0;
    //素数
    for(i=2;i<=sqrt(a);i++){
        if(a%i==0)  return 0;
    }
    return 1;
}

int main()
{
    int m,n;
    scanf("%d%d",&m,&n);
    for(int i=m;i<=n;i++){
        if(judge(i)){
                printf("%d\n",i);
        }
    }
    return 0;
}

分金币

思路：骑士获得金币的过程可以分成多个阶段，第一个阶段有一枚，第二个阶段两枚……第n个阶段n枚；同时第n个阶段会持续n天。于是可以发现，第k个阶段可以得到k*k枚金币，最后获得的金币数目本质上就是多个m倍的由1到m的和的总和。同时还要注意：最后一个阶段可能没有过完就已经到了输入的天数。

#include<stdio.h>
//确定发多少金币每个阶段发多少天
int fact(int a){
    int sum = 0;
    for(int j=1;j<=a;j++){
        sum+=j*j;
    }
    return sum;
}
int sum(int a){
    return a*(1+a)/2;
}

int main() {
    int t,num=0,k;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        if(sum(i)>t){
            k=i-1;
            break;
        }
    }
    num = fact(k) + (k+1)*(t-sum(k));
    printf("%d",num);
    return 0;
}

PS：这里sum函数可以确定到第几个阶段从而确定最后一个阶段是否结束（例如，到第三个阶段共经历了1+2+3天，当输入4的时候，到达第四阶段但是第四阶段没有过完）

? ? ? ?所以当sum(i)返回的天数大于输入天数的时候则说明完整度过了(i-1)个阶段，余下(t-sum(i-1))天在第i个阶段。此时金币总数目便由前(i-1)个阶段以及第i个阶段中的部分天所获得的金币构成。

阶乘之和（暴力）

思路：当n足够大的时候，结果会超出unsigned long long int所能表示的范围，故需要使用高精度。如果不会用高精度，可以根据最多为1至50的阶乘之和，暴力打表。

简单粗暴的无脑代码如下：

#include<stdio.h>

char str[40][66]={
"43954713",
"522956313",
"6749977113",
"93928268313",
"1401602636313",
"22324392524313",
"378011820620313",
"6780385526348313",
"128425485935180313",
"2561327494111820313",
"53652269665821260313",
"1177652997443428940313",
"27029669736328405580313",
"647478071469567844940313",
"16158688114800553828940313",
"419450149241406189412940313",
"11308319599659758350180940313",
"316196664211373618851684940313",
"9157958657951075573395300940313",
"274410818470142134209703780940313",
"8497249472648064951935266660940313",
"271628086406341595119153278820940313",
"8954945705218228090637347680100940313",
"304187744744822368938255957323620940313",
"10637335711130967298604907294846820940313",
"382630662501032184766604355445682020940313",
"14146383753727377231082583937026584420940313",
"537169001220328488991089808037100875620940313",
"20935051082417771847631371547939998232420940313",
"836850334330315506193242641144055892504420940313",
"34289376947494122614363304694584807557656420940313",
"1439295494700374021157505910939096377494040420940313",
"61854558558074209658512637979453093884758552420940313",
"2720126133346522977702138448994068984204397080420940313",
"122342346998826717539665299944651784048588130840420940313",
"5624964506810915667389970728744906677010239883800420940313",
"264248206017979096310354325882356886646207872272920420940313",
"12678163798554051767172643373255731925167694226950680420940313",
"620960027832821612639424806694551108812720525606160920420940313",
"31035053229546199656252032972759319953190362094566672920420940313"
};
int main()
{
    int n,factor=1,sum=0;
    while(scanf("%d",&n)!=EOF&&n)
    if(n<=10){
        for(int i=1;i<=n;i++){
            factor=1;
            for(int j=1;j<=i;j++)
                factor*=j;
            sum+=factor;
        }
        printf("%d\n",sum);
        sum = 0;
    }
    else if(n>10)   printf("%s\n",str[n-11]);
    return 0;
}

PS：可以看到，当n足够小的时候，结果甚至可以用int表示，这个时候可以用n<=10部分的代码计算结果。（当用unsigned long long int 来输出的时候，n可以比使用int时的n再大一点

数字方形与三角

思路：方形只需要两个循环分别控制行和列输出即可；三角则需要先控制“? ”的输出，再进行数字的输出，可以根据数字与行号的关系进行控制。注意两部分之间有空行。不谈~

参考码：

#include<stdio.h>

void print_square(int n){
    int kase = 1;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            printf("%02d",kase);
            kase++;
        }
        printf("\n");
    }
}
void print_triangle(int n){
    int kase = 1;
    for(int i=1;i<=n;i++){
        for(int j=n-i;j>0;j--){
            printf("  ");
        }
        for(int j=n-i;j<n;j++){
            printf("%02d",kase);
            kase++;
        }
        printf("\n");
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    print_square(n);
    printf("\n");
    print_triangle(n);
    return 0;
}

存钱💴

思路：用循环模拟每个月钱的收入和支出：先收入300元，然后根据这个月的预算判断是否存钱，得到存钱后手中剩下的钱和存起来的钱；年底存款再涨20%。输出只有预算大于手中钱数的月份和存款数两种情况。

这个码有点捞：

#include<stdio.h>

int main()
{
    int n,t = 0,kase = 0,m,deposit = 0;
    for(int i=1;i<=12;i++){
        t+=300;
        scanf("%d",&n);
        if(kase)    continue;
        if(t < n){
            kase = i;
        }
        if(t-n>=100){
            m = (t-n)/100;
            deposit = deposit + m;
            t = t - 100*m;
        }
        t-=n;
        //printf("%d月还剩下：%d元 %d存了：%d元\n",i,t,m,deposit);
    }
    if(kase)   printf("-%d",kase);
    else if(!kase)  printf("%d",t+600*deposit/5);
    return 0;
}

PS：存起来的钱只能为整百，再乘以1.2以后还是整数，所以可以用int类型。最后的钱数为剩下的钱加上存起来的钱数。

计数问题

思路：对于每个数只需要依次求出每一位上的数字，然后让对应数字出现次数加一，最后输出所需要的数字的出现次数就可以。可以通过取余来从后往前依次获得每一位上的数字。

这样，每次取余计数之后再将数字除以10，依次循环，从而取到每一位上的数字。于是，菜鸟稍加思考，如是码道：

#include<stdio.h>

void print_count(int num,int n){
    int a[10]={0};
    int i,j,k;
    for(i=0;i<=num;i++){
            k=i;
        for(j=0;;j++){
            if(k==0)    break;
            else{
                a[k%10]++;
                k/=10;
            }
        }
    }
    printf("%d",a[n]);
}

int main()
{
    int i,n,num;
    scanf("%d%d",&num,&n);
    print_count(num,n);
    return 0;
}

PS：开辟一个数组，让它从0到9号分别对应数字0到9，这样可以直接实现出现次数加一的操作。

文章来源:https://blog.csdn.net/2302_81103640/article/details/135309559
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