1 条题解

  • 3
    @ 2026-2-26 17:52:07

    思路

    1. 先依据美味度排序
    2. 如果前两项口味不同,直接输出前两项的美味度之和
    3. 如果前两项的口味相同,先去比较其它不同口味的,如果按顺序第二个口味中,最美味的那个(称为A),与所有冰激凌最甜的那个(称为B),之和比第二美味的冰激淋(称为C)与A的和还大,输出A和B的甜度之和,否则输出B和C的甜度之和

    加密代码

    偏移量:78,额外使用x模式

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    解密代码

    #define ZAYq1URqHRwS7qAv082 a
    #define rUcWATN5o9L_RbLohv6 b
    #define j9YcgzjhMHaBQ begin
    #define UCOGjSPMa9WS_sVVEDB bool
    #define Y5Ozh0jGtcdjKk break
    #define bM9jCUglv4bH8GX cin
    #define qU4vZqjZI4E cmp
    #define NVntjB6iO30iuiWJE cout
    #define pm8VxjRHuL4nks df
    #define xQ10rk9BpoSkAWpD else
    #define tHVTXLW0WnKaFCb6iG1v end
    #define bRej_exfXKgHXjP false
    #define o4kVCNMscyHYUft1gI for
    #define WM60zMzeMZS1eL i
    #define Wfg2aCwTPU icecream
    #define W6Ekoo4zwG if
    #define SHDxMJMCO int
    #define iB8tp3911KbVJl8Bz ios
    #define lAt2xko4WC0GNfFl ll
    #define xhEjNLW2KC_i long
    #define Rnb2PVqqctjILPZ8N main
    #define y25ZaTyH max
    #define hq55zRaRAbMn5O3a4 n
    #define nsweGl7H namespace
    #define T_v3Yg5a5Fq3dPl return
    #define ML_ZL445 sort
    #define kvk7p4Ds std
    #define ZRfkOo9eLFK5 struct
    #define wQEv0cPuCshO_Mlb style
    #define IpaA76tS9ELtPl sweet
    #define K3p1Nez1 sync_with_stdio
    #define VEI0KIjYGR71T tie
    #define XSAfvxPGR using
    #define kpxPwunrXdvm8 v
    #define UzZfIwOFfd4fy vector
    
    #include<bits/stdc++.h>
    #define ll long long
    XSAfvxPGR nsweGl7H kvk7p4Ds;
    
    ZRfkOo9eLFK5 Wfg2aCwTPU {
        lAt2xko4WC0GNfFl wQEv0cPuCshO_Mlb,IpaA76tS9ELtPl;
    };
    
    UCOGjSPMa9WS_sVVEDB qU4vZqjZI4E(Wfg2aCwTPU ZAYq1URqHRwS7qAv082,Wfg2aCwTPU rUcWATN5o9L_RbLohv6) {
        T_v3Yg5a5Fq3dPl ZAYq1URqHRwS7qAv082.IpaA76tS9ELtPl>rUcWATN5o9L_RbLohv6.IpaA76tS9ELtPl;
    }
    
    SHDxMJMCO Rnb2PVqqctjILPZ8N() {
        iB8tp3911KbVJl8Bz::K3p1Nez1(bRej_exfXKgHXjP);
        bM9jCUglv4bH8GX.VEI0KIjYGR71T(0);
        NVntjB6iO30iuiWJE.VEI0KIjYGR71T(0);
    
        lAt2xko4WC0GNfFl hq55zRaRAbMn5O3a4;
        bM9jCUglv4bH8GX>>hq55zRaRAbMn5O3a4;
        UzZfIwOFfd4fy<Wfg2aCwTPU> kpxPwunrXdvm8(hq55zRaRAbMn5O3a4+5);
        o4kVCNMscyHYUft1gI(lAt2xko4WC0GNfFl WM60zMzeMZS1eL=0;WM60zMzeMZS1eL<hq55zRaRAbMn5O3a4;WM60zMzeMZS1eL++) bM9jCUglv4bH8GX>>kpxPwunrXdvm8[WM60zMzeMZS1eL].wQEv0cPuCshO_Mlb>>kpxPwunrXdvm8[WM60zMzeMZS1eL].IpaA76tS9ELtPl;
        ML_ZL445(kpxPwunrXdvm8.j9YcgzjhMHaBQ(),kpxPwunrXdvm8.tHVTXLW0WnKaFCb6iG1v()-5,qU4vZqjZI4E);
        W6Ekoo4zwG(kpxPwunrXdvm8[0].wQEv0cPuCshO_Mlb!=kpxPwunrXdvm8[1].wQEv0cPuCshO_Mlb) NVntjB6iO30iuiWJE<<kpxPwunrXdvm8[0].IpaA76tS9ELtPl+kpxPwunrXdvm8[1].IpaA76tS9ELtPl;
        xQ10rk9BpoSkAWpD {
            lAt2xko4WC0GNfFl pm8VxjRHuL4nks=-1;
            o4kVCNMscyHYUft1gI(lAt2xko4WC0GNfFl WM60zMzeMZS1eL=2;WM60zMzeMZS1eL<hq55zRaRAbMn5O3a4;WM60zMzeMZS1eL++) {
                W6Ekoo4zwG(kpxPwunrXdvm8[0].wQEv0cPuCshO_Mlb!=kpxPwunrXdvm8[WM60zMzeMZS1eL].wQEv0cPuCshO_Mlb) {
                    pm8VxjRHuL4nks=kpxPwunrXdvm8[0].IpaA76tS9ELtPl+kpxPwunrXdvm8[WM60zMzeMZS1eL].IpaA76tS9ELtPl;
                    Y5Ozh0jGtcdjKk;
                }
            }
    
            NVntjB6iO30iuiWJE<<y25ZaTyH(kpxPwunrXdvm8[0].IpaA76tS9ELtPl+kpxPwunrXdvm8[1].IpaA76tS9ELtPl/2,pm8VxjRHuL4nks);
        }
    
        T_v3Yg5a5Fq3dPl 139;
    }
    
    • @ 2026-2-27 9:06:35

      最好不要看解密的,加密的代码并没有混淆

    • @ 2026-2-27 10:13:57

      感谢题解

  • 1

信息

ID
102
时间
1000ms
内存
256MiB
难度
3
标签
递交数
63
已通过
8
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