HDU 2021-2040

版权声明： https://blog.csdn.net/weixin_39778570/article/details/82529757

2021 发工资咯：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i,n) for(int i=1; i<=n; i++)
#define P pair<int, int>

int V[6] {1,2,5,10,50,100};
int solve(int x){
    int ans = 0;
    int now = x;
    for(int i=5; i>=0; i--){
        int t = now/V[i];
        ans += t;
        now -= t*V[i];
    }
    return ans;
}
int main(){
    int a,n,ans;
    while(scanf("%d", &n) && n!=0){
        ans = 0;
        REP(i,n){
            scanf("%d", &a);            
            ans += solve(a);
        }
        printf("%d\n", ans);
    }
    return 0;
}

2022海选女主角

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define REP(i,n) for(int i=0, i<n; i++)
#define REPP(i, n) for(int i=1;i<=n;i++)
#define P pair(int, int)

int n,m,a;
int main(){
    while(scanf("%d%d", &n,&m) != EOF){
        int x,y,ans=0,max=0;
        REPP(i,n){
            REPP(j,m){
                scanf("%d", &a);
                if(abs(a)>max){
                    max = abs(a);
                    ans = a;
                    x = i;
                    y = j;
                }
            }
        }
        printf("%d %d %d\n", x,y,ans);
    }
    return 0;
}

2023 求平均成绩

#include<bits/stdc++.h>
using namespace std;
#define LL long long 
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for(int i=1; i<=n; i++)
#define P pair(int, int)

int n, m, ans;
double now[100][100],s[100],k[100];
void out(){
    REP(i,n-1) printf("%.2f ", s[i]); 
    printf("%.2f\n", s[n-1]); 
    REP(i,m-1) printf("%.2f ", k[i]); 
    printf("%.2f\n", k[m-1]); 
    printf("%d\n\n", ans);
}
int main(){
    while(scanf("%d%d", &n,&m)!=EOF){
        memset(s,0,sizeof(s));
        memset(k,0,sizeof(k));
        REP(i,n){
            REP(j,m){
                scanf("%lf", &now[i][j]);
                s[i]+=now[i][j];
                k[j]+=now[i][j];
            }
        }
        REP(i,n) s[i] /= m;
        REP(i,m) k[i] /= n;
        ans = 0;
        REP(i,n){
            bool ok = 1;
            for(int j=0;j<m;j++){
                if(now[i][j]<k[j]){
                    ok = 0;
                    break;
                }
            }
            if(ok) ans++;
        }
        out();
    }
    return 0;
}

2024 C语言合法标识符

#include<bits/stdc++.h>
using namespace std;

#define LL long long
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i,n) for(int i=1; i<=n; i++)
#define P pair(int, int)

int t;
char c[105];
int main(){
    scanf("%d", &t);
    getchar(); // 获取缓冲区第一个字符 
    while(t--){
        gets(c); // 遇到回车结束，但回车不在字符串里 
        int n = strlen(c);  
        if(c[0]=='_' || isalpha(c[0])){
            bool ok = 1;
            REPP(i,n-1){
                if(isalnum(c[i]) || c[i]=='_') continue;
                else{
                    ok = 0;
                    break;
                }
            }
            if(ok) printf("yes\n");
            else   printf("no\n");
        }else printf("no\n"); 
    } 
    return 0;
}

2025 查找最大元素

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000 
#define REP(i,n) for(int i=0; i<n; i++) 
#define REPP(i,n) for(int i=1; i<=n; i++)
#define P pair(int, int)
#define LL long long

char p[MAXN];
int main(){
    while(scanf("%s", p) != EOF){
        int m = 0;
        char c = p[0];
        int len = strlen(p);
        REPP(i,len-1){
            if(p[i]>c){
                c = p[i];
                m = i;
            }
        }
        REP(i,len){
            if(p[i] == c) printf("%c(max)", c);
            else          printf("%c",p[i]);
        } 
        printf("\n");
    }

    return 0;
}

2026 首字母变大写

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000 
#define REP(i,n) for(int i=0; i<n; i++) 
#define REPP(i,n) for(int i=1; i<=n; i++)
#define P pair(int, int)
#define LL long long

char c[MAXN];
int main(){
    while(gets(c)){
        int len = strlen(c); 
        c[0] -= 32;
        REP(i, len){
            if(c[i]==' ') {
                c[i+1] -= 32;   
                printf(" ");
            }else{
                printf("%c", c[i]);
            }
        }
        printf("\n");
    }
}

2027 统计元音

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>

int A,E,I,O,U,t;
char s[105];
int main(){
    scanf("%d", &t);
    getchar(); // 回车
    while(t--){
        gets(s);
        A=E=I=O=U=0;
        int len = strlen(s);
        REP(i,len){
            if(s[i] == 'a') A++;
            else if(s[i] == 'e')E++;
            else if(s[i] == 'i')I++;
            else if(s[i] == 'o')O++;
            else if(s[i] == 'u')U++;
        }
        printf("a:%d\ne:%d\ni:%d\no:%d\nu:%d\n",A,E,I,O,U);
        if(t) printf("\n"); // 是每两个例子间空一行，最后不用 
    } 
    return 0;
}

2028 Lowest Common Multiple Plus
最小公倍数

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>

int n, a;
int gcd(LL a, LL b){
    if(a%b==0) return b;
    if(b==0)   return a; // 输入为0的时候要处理 
    else return gcd(b, a%b);
}
int main(){
    while(scanf("%d", &n)!=EOF){
        LL ans = 1;
        REP(i,n){
            scanf("%d", &a);
            ans = ans * a / gcd(ans,a);
        }
        printf("%d\n",  ans);
    }
    return 0;
} 

2029 Palindromes _easy version
判断回文

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>

int n;
char s[100000];
int main(){
    scanf("%d", &n);
    getchar();
    while(n--){
        gets(s);
        int len = strlen(s) - 1;
        bool ok = 1;
        for(int l=0,r=len; l<r; l++, r--){
            if(s[l]!=s[r]){
                ok = 0;
                break;
            }
        } 
        if(ok) printf("yes\n"); 
        else   printf("no\n");
    }
    return 0; 
} 

2030 汉字统计

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>

int n;
char a[100000];
int main(){
    scanf("%d", &n);
    getchar();
    while(n--){
        int ans = 0;
        gets(a);
        int len = strlen(a);
        REP(i,len){
            if(a[i]<0) ans++;  // 汉字的ASCILL码小于0 
        }
        printf("%d\n", ans/2); // 一个汉字占两个字节 
    }
    return 0;
}

2031 进制转换
10 转其他进制

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>
int a, r;
char R[] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
char ans[100000];
int main(){
    while(scanf("%d%d", &a,&r)!=EOF){
        int len = 0;
        int x = abs(a);
        while(x!=0){
            ans[len++] = R[x%r];
            x /= r;
        }
        if(a<0) printf("-");
        for(int i=len-1; i>=0; i--){
            printf("%c", ans[i]);
        }
        printf("\n");
    }
    return 0;
}

2032 杨辉三角

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>

int n;
LL a[40][40];
void init(){
    a[0][0] = 1;
    REPP(i, 40){
        a[i][0] = 1;
        REPP(j,i-1){
            a[i][j] = a[i-1][j-1] + a[i-1][j];  
        }
        a[i][i] = 1;
    }
}
void out(int n){
    for(int i=0; i<n; i++){
        printf("%lld",a[i][0]);
        for(int j=1; j<=i; j++){
            printf(" %lld", a[i][j]);
        }
        printf("\n");
    }
    printf("\n");
}
int main(){
    init();
    while(scanf("%d", &n)!=EOF){
        out(n);
    }
    return 0;
}

2033 人见人爱A+B

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>

int a1,a2,a3,b1,b2,b3,n;
int ans1, ans2, ans3;
int main(){
    scanf("%d", &n);
    while(n--){
        scanf("%d%d%d%d%d%d", &a1,&a2,&a3,&b1,&b2,&b3);
        ans1 = a1+b1;
        ans2 = a2+b2;
        ans3 = a3+b3;
        if(ans3>=60){ ans3-=60; ans2++;}
        if(ans2>=60){ ans2-=60; ans1++;}
        printf("%d %d %d\n", ans1, ans2, ans3);
    }
    return 0;
}

2034 人见人爱A-B
集合相减

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define LL long long
#define P pair<int, int>

int n,m,a;
set<int> ss;
int main(){
    while(scanf("%d%d", &n,&m) && (n!=0 || m!=0)){
        REP(i,n){
            scanf("%d", &a);
            ss.insert(a);   
        } 
        REP(i,m){
            scanf("%d", &a);
            ss.erase(a);    
        }
        if(ss.size()==0){
            printf("NULL"); 
        }else{
            for(set<int>::iterator it = ss.begin(); it!=ss.end(); it++) // !=, 平衡二叉树已经默认排序好了 
                printf("%d ", *it);
        }
        printf("\n");
        ss.clear(); // 记得清空 
    }

    return 0;
}

2035 人见人爱A^B
快速幂 要注意取后3位就是取余(同余)

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define ll long long
#define P pair<int, int>

ll mod_pow(ll x, ll n, ll mod=1){
    ll res = 1;
    while(n>0){
        if(n&1) res = res * x % mod;
        x = x*x % mod;
        n >>= 1;
    }
    return res;
}
ll a,b,ans;
int main(){
    while(scanf("%lld%lld", &a,&b) && (a!=0 || b!=0)){
    //  cout << mod_pow(a,b,1000) << endl;
        ans = mod_pow(a,b,1000); // 输出后3位 
        printf("%d\n", ans);
    }
    return 0;
} 

2036 改革春风吹满地
多边形面积 不用海伦公式的原因是因为计算量大 且精度丢失

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define ll long long
#define P pair<int, int>
/* 叉积 记得除2 */
struct F{
    double x,y;
};
double corss(const F &a, const F &b){
    return a.x*b.y - a.y*b.x;
}
int n;
F a, b, c;
int main(){
    while(scanf("%d", &n) && n!=0){
        double ans = 0.0;
        scanf("%lf%lf", &a.x,&a.y);
        b.x = a.x; b.y = a.y;
        for(int i=1; i<n; i++){
            scanf("%lf%lf", &c.x,&c.y);
            ans += corss(b,c); // 从第一个向量加乘到第n个向量 
            b = c;
        }
        ans += corss(c,a);  // 注意是闭合  是第n个向量乘第一个向量 
        printf("%.1f\n", abs(ans/2));
    }
    return 0;
} 

2037 今年暑假不AC
贪心

#include<bits/stdc++.h>
using namespace std;

#define MAXN 10000
#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for (int i=1; i<=n; i++)
#define ll long long
#define P pair<int, int>

int n,st,ed; 
vector<P> ver;
int main(){
    while(scanf("%d", &n) && n!=0){
        ver.clear();
        int ans = 0;
        for(int i=0; i<n; i++){
            scanf("%d%d", &st,&ed);
            ver.push_back(P(ed,st));
        }
        sort(ver.begin(),ver.end());
        ed = 0;
        for(int i=0; i<n; i++){
            if(ver[i].second>=ed){
                ans++;
                ed = ver[i].first;
            }
        }
        printf("%d\n", ans); 
    }

    return 0;
} 

2039 三角形
两边和要大于第三边 两边差小于第三边

#include<bits/stdc++.h>
using namespace std;

#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for(int i=1; i<=n; i++)
#define ll long long
#define P pair<int, int>

int n;
double a,b,c;
int main(){
    scanf("%d", &n);
    while(n--){
        scanf("%lf%lf%lf", &a, &b, &c); // 注意输入的不一定是整数,wa两次... 
        if(((a+b)>c && (a+c)>b && (b+c)>a) && (abs(a-b)<c && abs(a-c)<b && abs(b-c)<a)) printf("YES\n");
        else printf("NO\n");
    }       
    return 0;
} 

2040 亲和数

#include<bits/stdc++.h>
using namespace std;

#define REP(i,n) for(int i=0; i<n; i++)
#define REPP(i, n) for(int i=1; i<=n; i++)
#define ll long long
#define P pair<int, int>

ll qin(ll a){
    if(a==0 || a==1) return 0;
    ll res = 1;
    for(int i=2; i<=sqrt(a);i++)
        if(a%i==0) res += i+a/i;
    int q = (int)sqrt(a);   
    if(q*q==a) res -= q;
    return res;
}
int n,a,b; 
int main(){
    scanf("%d", &n);
    while(n--){
        scanf("%d%d", &a,&b);
        if(a==qin(b) && b==qin(a)) printf("YES\n");
        else printf("NO\n"); 
    }
    return 0;
}